Class 10 Mathematics Chapter 4 Question Answer | Quadratic Equations | English Medium

Class 10 Mathematics Chapter 4 Question Answer | Quadratic Equations | English Medium | ASSEB

Welcome dear learners to HSLC GURU! This page presents the complete English-medium question-answer guide for ASSEB (Assam State School Education Board) Class 10 Mathematics Chapter 4 — Quadratic Equations. Every textbook problem from Exercise 4.1, 4.2, 4.3 and 4.4 is solved step-by-step, along with a chapter summary, key formulas, additional MCQs and a glossary — all written in clean WordPress block format with KaTeX-rendered mathematics so you can study, revise and score full marks in your HSLC examination.

The chapter extends your earlier knowledge of polynomials and linear equations to second-degree equations in one variable. You will learn three powerful methods of solving them — factorisation (splitting the middle term), completing the square, and the celebrated quadratic formula — and how the discriminant reveals the nature of the roots without solving the equation. Real-life word problems on areas, ages, speeds, work, and numbers are translated into quadratic equations and solved.

Chapter Summary

Summary (English): A quadratic equation in the variable $x$ is an equation of the form $ax^2 + bx + c = 0$, where $a, b, c$ are real numbers and $a \ne 0$. Any equation that can be put in this standard form is also a quadratic equation. A real number $\alpha$ is called a root (or solution) of the quadratic equation $ax^2+bx+c=0$ if $a\alpha^2+b\alpha+c=0$. The zeros of the quadratic polynomial $ax^2+bx+c$ and the roots of the quadratic equation $ax^2+bx+c=0$ are the same. A quadratic equation always has at most two roots. Three methods are used for solving — factorisation by splitting the middle term, completing the square, and the quadratic formula. The expression $D = b^2 – 4ac$ is called the discriminant; it determines the nature of the roots: $D > 0$ gives two distinct real roots, $D = 0$ gives two equal real roots, and $D < 0$ gives no real roots.

Quadratic equations occur naturally in many situations — finding dimensions of plots, comparing speeds, calculating ages, working out the time taken by pipes to fill a tank, or solving geometry problems on right-angled triangles. The chapter ends with a powerful tool: the discriminant, which lets us decide without actually solving whether a real solution exists at all.

Standard Form of a Quadratic Equation

The general (standard) form of a quadratic equation in one variable $x$ is —

$$ax^2 + bx + c = 0, \quad a \ne 0$$

where $a$, $b$, $c$ are real numbers and $a$ is the coefficient of $x^2$ (called the leading coefficient), $b$ is the coefficient of $x$, and $c$ is the constant term. The condition $a \ne 0$ is essential — if $a = 0$, the equation reduces to a linear equation $bx + c = 0$.

Key Formulas at a Glance

Concept	Formula / Statement
Standard form	$ax^2 + bx + c = 0,\ a \ne 0$
Quadratic formula	$x = \dfrac{-b \pm \sqrt{b^2 – 4ac}}{2a}$
Discriminant	$D = b^2 – 4ac$
Two distinct real roots	$D > 0$
Two equal real roots	$D = 0$, root $= -\dfrac{b}{2a}$
No real roots	$D < 0$
Sum of roots	$\alpha + \beta = -\dfrac{b}{a}$
Product of roots	$\alpha \beta = \dfrac{c}{a}$
Equation from roots	$x^2 – (\alpha+\beta)x + \alpha\beta = 0$

The Quadratic Formula (Sridharacharya Formula)

The roots of the quadratic equation $ax^2+bx+c=0$ (with $a\ne 0$) are given by —

$$x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}$$

provided $b^2 – 4ac \ge 0$. This formula is named after the ancient Indian mathematician Sridharacharya (around 1025 CE) who is credited with the earliest known derivation. The two roots are usually denoted $\alpha$ (alpha) and $\beta$ (beta).

Three Methods of Solving

Method 1 — Factorisation (Splitting the Middle Term): Express $ax^2+bx+c$ as a product of two linear factors $(px+q)(rx+s)$ by splitting $b$ into two numbers whose sum is $b$ and product is $ac$. Then equate each factor to zero.

Method 2 — Completing the Square: Rewrite $ax^2+bx+c=0$ as $\left(x+\dfrac{b}{2a}\right)^2 = \dfrac{b^2-4ac}{4a^2}$ and take the square root.

Method 3 — Quadratic Formula: Apply $x = \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$ directly. This always works whenever $D \ge 0$.

Exercise 4.1 — Identifying Quadratic Equations & Forming Them

Q1. Check whether the following are quadratic equations:

(i) $(x+1)^2 = 2(x-3)$

Answer: LHS $= x^2 + 2x + 1$, RHS $= 2x – 6$. So $x^2 + 2x + 1 = 2x – 6$, which gives $x^2 + 7 = 0$. This is of the form $ax^2 + bx + c = 0$ with $a=1, b=0, c=7$. Hence it is a quadratic equation.

(ii) $x^2 – 2x = (-2)(3-x)$

Answer: RHS $= -6 + 2x$. So $x^2 – 2x = -6 + 2x \Rightarrow x^2 – 4x + 6 = 0$. This is a quadratic equation with $a=1, b=-4, c=6$.

(iii) $(x-2)(x+1) = (x-1)(x+3)$

Answer: LHS $= x^2 – x – 2$, RHS $= x^2 + 2x – 3$. Equating: $x^2 – x – 2 = x^2 + 2x – 3 \Rightarrow -3x + 1 = 0$. This is a linear equation, hence not a quadratic equation.

(iv) $(x-3)(2x+1) = x(x+5)$

Answer: LHS $= 2x^2 – 5x – 3$, RHS $= x^2 + 5x$. So $2x^2 – 5x – 3 – x^2 – 5x = 0 \Rightarrow x^2 – 10x – 3 = 0$. Yes, it is a quadratic equation.

(v) $(2x-1)(x-3) = (x+5)(x-1)$

Answer: LHS $= 2x^2 – 7x + 3$, RHS $= x^2 + 4x – 5$. Difference: $x^2 – 11x + 8 = 0$. Yes, it is a quadratic equation.

(vi) $x^2 + 3x + 1 = (x-2)^2$

Answer: RHS $= x^2 – 4x + 4$. So $x^2 + 3x + 1 – x^2 + 4x – 4 = 0 \Rightarrow 7x – 3 = 0$. This is linear, not a quadratic equation.

(vii) $(x+2)^3 = 2x(x^2-1)$

Answer: LHS $= x^3 + 6x^2 + 12x + 8$, RHS $= 2x^3 – 2x$. So $x^3 + 6x^2 + 12x + 8 – 2x^3 + 2x = 0 \Rightarrow -x^3 + 6x^2 + 14x + 8 = 0$, i.e. $x^3 – 6x^2 – 14x – 8 = 0$. This is a cubic, hence not quadratic.

(viii) $x^3 – 4x^2 – x + 1 = (x-2)^3$

Answer: RHS $= x^3 – 6x^2 + 12x – 8$. Difference: $x^3 – 4x^2 – x + 1 – x^3 + 6x^2 – 12x + 8 = 0 \Rightarrow 2x^2 – 13x + 9 = 0$. Yes, it is a quadratic equation.

Q2. Represent the following situations in the form of quadratic equations:

(i) The area of a rectangular plot is $528\,\text{m}^2$. The length of the plot (in metres) is one more than twice its breadth. We need to find the length and breadth of the plot.

Answer: Let the breadth be $x$ m. Then length $= (2x+1)$ m. Area $= x(2x+1) = 528 \Rightarrow 2x^2 + x – 528 = 0$. This is the required quadratic equation.

(ii) The product of two consecutive positive integers is 306. We need to find the integers.

Answer: Let the smaller integer be $x$. Then the next is $(x+1)$. So $x(x+1) = 306 \Rightarrow x^2 + x – 306 = 0$.

(iii) Rohan’s mother is 26 years older than him. The product of their ages (in years) 3 years from now will be 360. We would like to find Rohan’s present age.

Answer: Let Rohan’s present age be $x$ years. Then mother’s present age $= (x+26)$ years. After 3 years: Rohan $=(x+3)$, mother $=(x+29)$. Product $(x+3)(x+29)=360 \Rightarrow x^2 + 32x + 87 = 360 \Rightarrow x^2 + 32x – 273 = 0$.

(iv) A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km/h less, then it would have taken 3 hours more to cover the same distance. We need to find the speed of the train.

Answer: Let the uniform speed be $x$ km/h. Time taken $= \dfrac{480}{x}$ h. With reduced speed $(x-8)$, time $= \dfrac{480}{x-8}$ h. Given $\dfrac{480}{x-8} – \dfrac{480}{x} = 3$. Multiplying through by $x(x-8)$: $480x – 480(x-8) = 3x(x-8) \Rightarrow 3840 = 3x^2 – 24x \Rightarrow x^2 – 8x – 1280 = 0$. This is the required quadratic equation.

Exercise 4.2 — Solving by Factorisation

Q1. Find the roots of the following quadratic equations by factorisation:

(i) $x^2 – 3x – 10 = 0$

Answer: Split $-3x$ as $-5x + 2x$ (since $-5 \times 2 = -10$ and $-5 + 2 = -3$). So $x^2 – 5x + 2x – 10 = 0 \Rightarrow x(x-5) + 2(x-5) = 0 \Rightarrow (x-5)(x+2) = 0$. Hence $x = 5$ or $x = -2$.

(ii) $2x^2 + x – 6 = 0$

Answer: Product $= 2 \times (-6) = -12$. We need two numbers whose product is $-12$ and sum is $1$: choose $4$ and $-3$. So $2x^2 + 4x – 3x – 6 = 0 \Rightarrow 2x(x+2) – 3(x+2) = 0 \Rightarrow (x+2)(2x-3)=0$. Hence $x=-2$ or $x=\dfrac{3}{2}$.

(iii) $\sqrt{2}\,x^2 + 7x + 5\sqrt{2} = 0$

Answer: Product $= \sqrt{2} \cdot 5\sqrt{2} = 10$. Need two numbers with product $10$ and sum $7$: choose $5$ and $2$. So $\sqrt{2}\,x^2 + 5x + 2x + 5\sqrt{2} = 0 \Rightarrow x(\sqrt{2}\,x + 5) + \sqrt{2}(\sqrt{2}\,x + 5) = 0 \Rightarrow (\sqrt{2}\,x + 5)(x + \sqrt{2}) = 0$. Hence $x = -\dfrac{5}{\sqrt{2}} = -\dfrac{5\sqrt{2}}{2}$ or $x = -\sqrt{2}$.

(iv) $2x^2 – x + \dfrac{1}{8} = 0$

Answer: Multiply by 8: $16x^2 – 8x + 1 = 0$. This is $(4x-1)^2 = 0$. Hence $x = \dfrac{1}{4}, \dfrac{1}{4}$ (equal real roots).

(v) $100x^2 – 20x + 1 = 0$

Answer: Note $100x^2 – 20x + 1 = (10x – 1)^2 = 0$. Hence $x = \dfrac{1}{10}, \dfrac{1}{10}$.

Q2. Solve the problems given:

(i) John and Jivanti together have 45 marbles. Both of them lost 5 marbles each, and the product of the number of marbles they now have is 124. We would like to find out how many marbles they had to start with.

Answer: Let John have $x$ marbles. Then Jivanti has $(45 – x)$. After losing 5 each: John $(x-5)$, Jivanti $(40-x)$. Product: $(x-5)(40-x) = 124 \Rightarrow 40x – x^2 – 200 + 5x = 124 \Rightarrow -x^2 + 45x – 200 = 124 \Rightarrow x^2 – 45x + 324 = 0$. Splitting $-45 = -36 – 9$ (product $324$): $(x-36)(x-9) = 0$. So $x = 36$ or $x = 9$. Hence the marbles were either $(36, 9)$ or $(9, 36)$.

(ii) A cottage industry produces a certain number of toys in a day. The cost of production of each toy (in rupees) was found to be 55 minus the number of toys produced in a day. On a particular day, the total cost of production was Rs 750. We would like to find out the number of toys produced on that day.

Answer: Let the number of toys produced be $x$. Cost per toy $= (55 – x)$ rupees. Total cost: $x(55 – x) = 750 \Rightarrow 55x – x^2 = 750 \Rightarrow x^2 – 55x + 750 = 0$. Splitting $-55 = -25 – 30$ (product $750$): $(x-25)(x-30) = 0$. So $x = 25$ or $x = 30$. Hence either 25 or 30 toys were produced.

Q3. Find two numbers whose sum is 27 and product is 182.

Answer: Let one number be $x$. Then the other is $(27 – x)$. Product: $x(27-x) = 182 \Rightarrow x^2 – 27x + 182 = 0$. Splitting: $(x-13)(x-14) = 0$. So $x = 13$ or $x = 14$. The numbers are 13 and 14.

Q4. Find two consecutive positive integers, sum of whose squares is 365.

Answer: Let the smaller integer be $x$. Then the next is $(x+1)$. So $x^2 + (x+1)^2 = 365 \Rightarrow 2x^2 + 2x + 1 = 365 \Rightarrow 2x^2 + 2x – 364 = 0 \Rightarrow x^2 + x – 182 = 0$. Splitting: $(x-13)(x+14) = 0$. Since $x$ is positive, $x = 13$. Hence the integers are 13 and 14.

Q5. The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.

Answer: Let the base be $x$ cm. Then altitude $= (x-7)$ cm. By Pythagoras, $x^2 + (x-7)^2 = 13^2 \Rightarrow x^2 + x^2 – 14x + 49 = 169 \Rightarrow 2x^2 – 14x – 120 = 0 \Rightarrow x^2 – 7x – 60 = 0$. Splitting: $(x-12)(x+5) = 0$. Since $x > 0$, $x = 12$. Hence base $= 12$ cm and altitude $= 5$ cm.

Q6. A cottage industry produces articles. The cost of production of each article (in rupees) is 3 more than twice the number of articles produced on that day. If the total cost of production on that day was Rs 90, find the number of articles produced and the cost of each article.

Answer: Let the number of articles be $x$. Cost per article $= (2x + 3)$ rupees. Total: $x(2x+3) = 90 \Rightarrow 2x^2 + 3x – 90 = 0$. Splitting $3 = 15 – 12$ (product $-180$): $2x^2 + 15x – 12x – 90 = 0 \Rightarrow x(2x+15) – 6(2x+15) = 0 \Rightarrow (2x+15)(x-6) = 0$. Since $x > 0$, $x = 6$. Hence 6 articles were produced and cost per article $= 2(6)+3 = $ Rs 15.

Exercise 4.3 — Completing the Square & Quadratic Formula

Q1. Find the roots of the following quadratic equations, if they exist, by the method of completing the square:

(i) $2x^2 – 7x + 3 = 0$

Answer: Divide by 2: $x^2 – \dfrac{7}{2}x + \dfrac{3}{2} = 0 \Rightarrow x^2 – \dfrac{7}{2}x = -\dfrac{3}{2}$. Add $\left(\dfrac{7}{4}\right)^2 = \dfrac{49}{16}$ to both sides: $\left(x – \dfrac{7}{4}\right)^2 = \dfrac{49}{16} – \dfrac{3}{2} = \dfrac{49 – 24}{16} = \dfrac{25}{16}$. Taking square root: $x – \dfrac{7}{4} = \pm \dfrac{5}{4}$. So $x = \dfrac{7}{4} + \dfrac{5}{4} = 3$ or $x = \dfrac{7}{4} – \dfrac{5}{4} = \dfrac{1}{2}$. Hence $x = 3, \dfrac{1}{2}$.

(ii) $2x^2 + x – 4 = 0$

Answer: Divide by 2: $x^2 + \dfrac{1}{2}x = 2$. Add $\left(\dfrac{1}{4}\right)^2 = \dfrac{1}{16}$ to both sides: $\left(x + \dfrac{1}{4}\right)^2 = 2 + \dfrac{1}{16} = \dfrac{33}{16}$. So $x + \dfrac{1}{4} = \pm \dfrac{\sqrt{33}}{4}$. Hence $x = \dfrac{-1 \pm \sqrt{33}}{4}$.

(iii) $4x^2 + 4\sqrt{3}\,x + 3 = 0$

Answer: Divide by 4: $x^2 + \sqrt{3}\,x + \dfrac{3}{4} = 0$. Note this is $\left(x + \dfrac{\sqrt{3}}{2}\right)^2 = 0$. Hence $x = -\dfrac{\sqrt{3}}{2}, -\dfrac{\sqrt{3}}{2}$ (equal real roots).

(iv) $2x^2 + x + 4 = 0$

Answer: Divide by 2: $x^2 + \dfrac{1}{2}x + 2 = 0$. So $x^2 + \dfrac{1}{2}x = -2$. Add $\dfrac{1}{16}$: $\left(x + \dfrac{1}{4}\right)^2 = -2 + \dfrac{1}{16} = -\dfrac{31}{16}$. Since RHS is negative, the square of a real number cannot be negative. Hence the equation has no real roots.

Q2. Find the roots of the quadratic equations given in Q1 above by applying the quadratic formula.

Answer: Apply $x = \dfrac{-b \pm \sqrt{b^2 – 4ac}}{2a}$ to each:

(i) $2x^2-7x+3=0$: $a=2, b=-7, c=3$. $D = 49 – 24 = 25 > 0$. $x = \dfrac{7 \pm 5}{4}$. So $x = 3, \dfrac{1}{2}$.
(ii) $2x^2+x-4=0$: $a=2, b=1, c=-4$. $D = 1 + 32 = 33$. $x = \dfrac{-1 \pm \sqrt{33}}{4}$.
(iii) $4x^2+4\sqrt{3}\,x+3=0$: $a=4, b=4\sqrt{3}, c=3$. $D = 48 – 48 = 0$. $x = \dfrac{-4\sqrt{3}}{8} = -\dfrac{\sqrt{3}}{2}$ (repeated).
(iv) $2x^2+x+4=0$: $D = 1 – 32 = -31 < 0$. No real roots.

Q3. Find the roots of the following equations:

(i) $x – \dfrac{1}{x} = 3$, $x \ne 0$

Answer: Multiply both sides by $x$: $x^2 – 1 = 3x \Rightarrow x^2 – 3x – 1 = 0$. Apply formula: $a=1, b=-3, c=-1$, $D = 9 + 4 = 13$. $x = \dfrac{3 \pm \sqrt{13}}{2}$.

(ii) $\dfrac{1}{x+4} – \dfrac{1}{x-7} = \dfrac{11}{30}$, $x \ne -4, 7$

Answer: LHS $= \dfrac{(x-7) – (x+4)}{(x+4)(x-7)} = \dfrac{-11}{(x+4)(x-7)}$. So $\dfrac{-11}{(x+4)(x-7)} = \dfrac{11}{30}$. Cross-multiplying: $-30 = (x+4)(x-7) \Rightarrow x^2 – 3x – 28 = -30 \Rightarrow x^2 – 3x + 2 = 0$. Splitting: $(x-1)(x-2) = 0$. So $x = 1$ or $x = 2$.

Q4. The sum of the reciprocals of Rehman’s ages (in years) 3 years ago and 5 years from now is $\dfrac{1}{3}$. Find his present age.

Answer: Let Rehman’s present age be $x$ years. Then $\dfrac{1}{x-3} + \dfrac{1}{x+5} = \dfrac{1}{3}$. So $\dfrac{(x+5)+(x-3)}{(x-3)(x+5)} = \dfrac{1}{3} \Rightarrow \dfrac{2x+2}{x^2+2x-15} = \dfrac{1}{3}$. Cross-multiplying: $3(2x+2) = x^2 + 2x – 15 \Rightarrow 6x + 6 = x^2 + 2x – 15 \Rightarrow x^2 – 4x – 21 = 0$. Splitting: $(x-7)(x+3) = 0$. Since age is positive, $x = 7$. Rehman’s present age is 7 years.

Q5. In a class test, the sum of Shefali’s marks in Mathematics and English is 30. Had she got 2 marks more in Mathematics and 3 marks less in English, the product of their marks would have been 210. Find her marks in the two subjects.

Answer: Let Shefali’s marks in Mathematics be $x$. Then in English $= (30 – x)$. New marks: Maths $(x+2)$, English $(30-x-3) = (27-x)$. Product: $(x+2)(27-x) = 210 \Rightarrow 27x – x^2 + 54 – 2x = 210 \Rightarrow -x^2 + 25x + 54 – 210 = 0 \Rightarrow x^2 – 25x + 156 = 0$. Splitting $-25 = -12 – 13$ (product $156$): $(x-12)(x-13) = 0$. So $x = 12$ or $x = 13$. Hence (Maths, English) = $(12, 18)$ or $(13, 17)$.

Q6. The diagonal of a rectangular field is 60 metres more than the shorter side. If the longer side is 30 metres more than the shorter side, find the sides of the field.

Answer: Let the shorter side be $x$ m. Then longer side $= (x+30)$ m and diagonal $= (x+60)$ m. By Pythagoras: $x^2 + (x+30)^2 = (x+60)^2 \Rightarrow x^2 + x^2 + 60x + 900 = x^2 + 120x + 3600 \Rightarrow x^2 – 60x – 2700 = 0$. Apply formula: $D = 3600 + 10800 = 14400$, $\sqrt{D} = 120$. $x = \dfrac{60 \pm 120}{2}$. Taking positive value, $x = 90$. Hence shorter side $= 90$ m, longer side $= 120$ m, diagonal $= 150$ m.

Q7. The difference of squares of two numbers is 180. The square of the smaller number is 8 times the larger number. Find the two numbers.

Answer: Let the larger number be $x$ and smaller be $y$. Then $x^2 – y^2 = 180$ and $y^2 = 8x$. Substituting: $x^2 – 8x = 180 \Rightarrow x^2 – 8x – 180 = 0$. Splitting $-8 = -18 + 10$ (product $-180$): $(x-18)(x+10) = 0$. Since $y^2 = 8x \ge 0$, $x \ge 0$, so $x = 18$. Then $y^2 = 8(18) = 144$, $y = \pm 12$. The numbers are 18 and 12 (or 18 and $-12$).

Q8. A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, it would have taken 1 hour less for the same journey. Find the speed of the train.

Answer: Let speed $= x$ km/h. Time $=\dfrac{360}{x}$ h. With increased speed: $\dfrac{360}{x+5}$. Given $\dfrac{360}{x} – \dfrac{360}{x+5} = 1$. Multiply by $x(x+5)$: $360(x+5) – 360x = x(x+5) \Rightarrow 1800 = x^2 + 5x \Rightarrow x^2 + 5x – 1800 = 0$. $D = 25 + 7200 = 7225$, $\sqrt{D} = 85$. $x = \dfrac{-5 + 85}{2} = 40$. Hence speed of train $= 40$ km/h.

Q9. Two water taps together can fill a tank in $9\dfrac{3}{8}$ hours. The tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.

Answer: Let the smaller tap take $x$ h to fill alone. Then the larger tap takes $(x-10)$ h. Together time $= \dfrac{75}{8}$ h. So $\dfrac{1}{x} + \dfrac{1}{x-10} = \dfrac{8}{75}$. LHS $= \dfrac{(x-10)+x}{x(x-10)} = \dfrac{2x-10}{x^2-10x}$. So $\dfrac{2x-10}{x^2-10x} = \dfrac{8}{75}$. Cross-multiplying: $75(2x-10) = 8(x^2-10x) \Rightarrow 150x – 750 = 8x^2 – 80x \Rightarrow 8x^2 – 230x + 750 = 0 \Rightarrow 4x^2 – 115x + 375 = 0$. Apply formula: $D = 13225 – 6000 = 7225$, $\sqrt{D} = 85$. $x = \dfrac{115 \pm 85}{8}$. So $x = 25$ or $x = \dfrac{30}{8} = \dfrac{15}{4}$. Since $x > 10$, $x = 25$. Hence the smaller tap takes 25 hours and the larger tap takes 15 hours.

Q10. An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bangalore (without taking into consideration the time they stop at intermediate stations). If the average speed of the express train is 11 km/h more than that of the passenger train, find the average speed of the two trains.

Answer: Let passenger train speed $= x$ km/h. Then express train speed $= (x+11)$ km/h. Given $\dfrac{132}{x} – \dfrac{132}{x+11} = 1$. Multiplying by $x(x+11)$: $132(x+11) – 132x = x(x+11) \Rightarrow 1452 = x^2 + 11x \Rightarrow x^2 + 11x – 1452 = 0$. $D = 121 + 5808 = 5929$, $\sqrt{D} = 77$. $x = \dfrac{-11 + 77}{2} = 33$. Hence passenger train speed $= 33$ km/h, express train speed $= 44$ km/h.

Q11. Sum of the areas of two squares is $468\,\text{m}^2$. If the difference of their perimeters is 24 m, find the sides of the two squares.

Answer: Let the sides be $x$ m and $y$ m with $x > y$. Then $4x – 4y = 24 \Rightarrow x – y = 6 \Rightarrow x = y + 6$. Also $x^2 + y^2 = 468$. Substituting: $(y+6)^2 + y^2 = 468 \Rightarrow 2y^2 + 12y + 36 = 468 \Rightarrow 2y^2 + 12y – 432 = 0 \Rightarrow y^2 + 6y – 216 = 0$. Splitting $6 = 18 – 12$ (product $-216$): $(y+18)(y-12) = 0$. Since $y > 0$, $y = 12$. So $x = 18$. The sides are 18 m and 12 m.

Exercise 4.4 — Nature of Roots (Discriminant)

Q1. Find the nature of the roots of the following quadratic equations. If the real roots exist, find them:

(i) $2x^2 – 3x + 5 = 0$

Answer: $a=2, b=-3, c=5$. $D = b^2 – 4ac = 9 – 40 = -31 < 0$. Hence the equation has no real roots.

(ii) $3x^2 – 4\sqrt{3}\,x + 4 = 0$

Answer: $a=3, b=-4\sqrt{3}, c=4$. $D = 48 – 48 = 0$. Hence two equal real roots. $x = -\dfrac{b}{2a} = \dfrac{4\sqrt{3}}{6} = \dfrac{2\sqrt{3}}{3} = \dfrac{2}{\sqrt{3}}$. So $x = \dfrac{2}{\sqrt{3}}, \dfrac{2}{\sqrt{3}}$.

(iii) $2x^2 – 6x + 3 = 0$

Answer: $a=2, b=-6, c=3$. $D = 36 – 24 = 12 > 0$. Hence two distinct real roots. $x = \dfrac{6 \pm \sqrt{12}}{4} = \dfrac{6 \pm 2\sqrt{3}}{4} = \dfrac{3 \pm \sqrt{3}}{2}$.

Q2. Find the values of $k$ for each of the following quadratic equations, so that they have two equal roots:

(i) $2x^2 + kx + 3 = 0$

Answer: For equal roots, $D = 0$. Here $a=2, b=k, c=3$. So $k^2 – 4(2)(3) = 0 \Rightarrow k^2 = 24 \Rightarrow k = \pm 2\sqrt{6}$.

(ii) $kx(x-2) + 6 = 0$

Answer: Expand: $kx^2 – 2kx + 6 = 0$. So $a=k, b=-2k, c=6$. For equal roots $D=0$: $4k^2 – 24k = 0 \Rightarrow 4k(k-6)=0$. So $k=0$ or $k=6$. But $k = 0$ would make the equation non-quadratic. Hence $k = 6$.

Q3. Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is $800\,\text{m}^2$? If so, find its length and breadth.

Answer: Let the breadth be $x$ m. Then length $= 2x$ m. Area: $2x^2 = 800 \Rightarrow x^2 = 400 \Rightarrow x = 20$ (positive). $D = 0 + 4(2)(800) = 6400 > 0$. Yes, design is possible. Length = 40 m, breadth = 20 m.

Q4. Is the following situation possible? If so, determine their present ages. The sum of the ages of two friends is 20 years. Four years ago, the product of their ages in years was 48.

Answer: Let the present age of one friend be $x$ years. Then the other is $(20-x)$ years. Four years ago: $(x-4)(16-x) = 48 \Rightarrow 16x – x^2 – 64 + 4x = 48 \Rightarrow -x^2 + 20x – 112 = 0 \Rightarrow x^2 – 20x + 112 = 0$. $D = 400 – 448 = -48 < 0$. Hence no real roots; the situation is not possible.

Q5. Is it possible to design a rectangular park of perimeter 80 m and area $400\,\text{m}^2$? If so, find its length and breadth.

Answer: Let length $= l$ m and breadth $= b$ m. Then $2(l+b) = 80 \Rightarrow l + b = 40$. Area $= lb = 400$. So $l$ and $b$ are roots of $x^2 – 40x + 400 = 0$. $D = 1600 – 1600 = 0$. Hence equal real roots. $x = \dfrac{40}{2} = 20$. So $l = b = 20$ m. The park is a square of side 20 m; design is possible.

Additional Practice — MCQs

1. Which of the following is NOT a quadratic equation?

(a) $x^2 + 3x – 5 = 0$
(b) $x^3 + x^2 + 2 = 0$
(c) $3 + x + x^2 = 0$
(d) $x^2 – 9 = 0$

Answer: (b) $x^3 + x^2 + 2 = 0$ — degree 3, not quadratic.

2. The degree of a quadratic equation is — (a) 0 (b) 1 (c) 2 (d) 3

Answer: (c) 2.

3. The discriminant of $ax^2 + bx + c = 0$ is — (a) $b^2 + 4ac$ (b) $b^2 – 4ac$ (c) $4ac – b^2$ (d) $-b^2 – 4ac$

Answer: (b) $b^2 – 4ac$.

4. The roots of $7x^2 + x – 1 = 0$ are — (a) Real and distinct (b) Real and equal (c) Not real (d) None

Answer: (a) Real and distinct, since $D = 1 + 28 = 29 > 0$.

5. Two consecutive natural numbers whose sum of squares is 313 are — (a) 12, 13 (b) 13, 14 (c) 11, 12 (d) 14, 15

Answer: (a) 12, 13 since $12^2 + 13^2 = 144 + 169 = 313$.

6. If $-5$ is a root of $2x^2 + px – 15 = 0$, then $p = ?$ (a) 3 (b) 5 (c) 7 (d) 1

Answer: (c) 7. Substituting: $2(25) + p(-5) – 15 = 0 \Rightarrow 50 – 5p – 15 = 0 \Rightarrow 5p = 35 \Rightarrow p = 7$.

7. If 2 is a root of $2x^2 + kx – 6 = 0$, then $k = ?$ (a) 1 (b) $-1$ (c) 2 (d) $-2$

Answer: (b) $-1$. Substituting: $8 + 2k – 6 = 0 \Rightarrow 2k = -2 \Rightarrow k = -1$.

8. The number of roots of $(x+2)^3 = x^3 – 4$ is — (a) 4 (b) 3 (c) 2 (d) 1

Answer: (c) 2. Expanding: $x^3 + 6x^2 + 12x + 8 = x^3 – 4 \Rightarrow 6x^2 + 12x + 12 = 0 \Rightarrow x^2 + 2x + 2 = 0$ (quadratic, hence at most 2 roots).

9. The roots of $x^2 + 7x + 12 = 0$ are — (a) $-3, -4$ (b) $3, 4$ (c) $-3, 4$ (d) $3, -4$

Answer: (a) $-3, -4$ since $(x+3)(x+4) = 0$.

10. If $\alpha$ and $\beta$ are roots of $x^2 – 5x + 6 = 0$, then $\alpha + \beta = ?$ (a) 5 (b) 6 (c) $-5$ (d) $-6$

Answer: (a) 5. Sum of roots $= -\dfrac{b}{a} = 5$.

11. The product of roots of $2x^2 – 7x + 3 = 0$ is — (a) $\dfrac{3}{2}$ (b) $-\dfrac{3}{2}$ (c) $\dfrac{7}{2}$ (d) $-\dfrac{7}{2}$

Answer: (a) $\dfrac{3}{2}$ since product $= \dfrac{c}{a} = \dfrac{3}{2}$.

12. For what value of $k$ does $kx^2 – 5x + k = 0$ have equal roots? (a) $\pm \dfrac{2}{5}$ (b) $\pm \dfrac{5}{2}$ (c) $\pm 5$ (d) $\pm 2$

Answer: (b) $\pm \dfrac{5}{2}$. $D = 25 – 4k^2 = 0 \Rightarrow k^2 = \dfrac{25}{4} \Rightarrow k = \pm \dfrac{5}{2}$.

Additional Practice — Short Answer

Q1. Find the value of $k$ for which $x = -2$ is a root of $kx^2 + x – 3 = 0$.

Answer: Substitute: $k(4) + (-2) – 3 = 0 \Rightarrow 4k = 5 \Rightarrow k = \dfrac{5}{4}$.

Q2. Form the quadratic equation whose roots are $3$ and $-2$.

Answer: Sum $= 3 + (-2) = 1$, product $= 3 \cdot (-2) = -6$. Equation: $x^2 – (\text{sum})x + \text{product} = 0 \Rightarrow x^2 – x – 6 = 0$.

Q3. Solve $9x^2 – 6x + 1 = 0$.

Answer: $9x^2 – 6x + 1 = (3x-1)^2 = 0$. So $x = \dfrac{1}{3}, \dfrac{1}{3}$.

Q4. Find the discriminant of $3x^2 – 2x + \dfrac{1}{3} = 0$ and hence find the nature of the roots.

Answer: $a = 3, b = -2, c = \dfrac{1}{3}$. $D = 4 – 4(3)\left(\dfrac{1}{3}\right) = 4 – 4 = 0$. Hence two equal real roots. $x = \dfrac{-b}{2a} = \dfrac{2}{6} = \dfrac{1}{3}$.

Q5. Solve by quadratic formula: $x^2 + 4x + 5 = 0$.

Answer: $D = 16 – 20 = -4 < 0$. Hence the equation has no real roots.

Q6. The sum of two numbers is 15 and the sum of their reciprocals is $\dfrac{3}{10}$. Find the numbers.

Answer: Let one number be $x$. Then the other is $(15 – x)$. Given $\dfrac{1}{x} + \dfrac{1}{15-x} = \dfrac{3}{10} \Rightarrow \dfrac{15}{x(15-x)} = \dfrac{3}{10} \Rightarrow 150 = 3x(15-x) \Rightarrow 50 = 15x – x^2 \Rightarrow x^2 – 15x + 50 = 0$. Splitting: $(x-5)(x-10) = 0$. So the numbers are 5 and 10.

Q7. The hypotenuse of a right triangle is 25 cm. The difference between the lengths of the other two sides is 5 cm. Find the lengths of these sides.

Answer: Let the smaller side be $x$ cm. Then the other is $(x+5)$ cm. By Pythagoras: $x^2 + (x+5)^2 = 625 \Rightarrow 2x^2 + 10x + 25 = 625 \Rightarrow x^2 + 5x – 300 = 0$. Splitting: $(x+20)(x-15) = 0$. Since $x > 0$, $x = 15$. Hence the sides are 15 cm and 20 cm.

Q8. A motor boat goes 30 km upstream and returns. Its total time is 4 hours 30 minutes. If the speed of the stream is 5 km/h, find the speed of the boat in still water.

Answer: Let speed in still water $= x$ km/h. Upstream speed $= (x-5)$, downstream $= (x+5)$. Total time $= 4.5$ h: $\dfrac{30}{x-5} + \dfrac{30}{x+5} = \dfrac{9}{2} \Rightarrow \dfrac{30(x+5) + 30(x-5)}{x^2 – 25} = \dfrac{9}{2} \Rightarrow \dfrac{60x}{x^2 – 25} = \dfrac{9}{2}$. Cross-multiplying: $120x = 9(x^2 – 25) \Rightarrow 9x^2 – 120x – 225 = 0 \Rightarrow 3x^2 – 40x – 75 = 0$. $D = 1600 + 900 = 2500$, $\sqrt{D} = 50$. $x = \dfrac{40 + 50}{6} = 15$. Hence speed of boat in still water $= $ 15 km/h.

Q9. The sum of squares of two consecutive odd positive integers is 290. Find the integers.

Answer: Let the smaller odd integer be $x$. Then the next is $(x+2)$. So $x^2 + (x+2)^2 = 290 \Rightarrow 2x^2 + 4x + 4 = 290 \Rightarrow x^2 + 2x – 143 = 0$. Splitting: $(x+13)(x-11) = 0$. Since $x > 0$, $x = 11$. Hence integers are 11 and 13.

Q10. Find the roots of $\dfrac{1}{x-2} + \dfrac{2}{x-1} = \dfrac{6}{x}$, $x \ne 0, 1, 2$.

Answer: LHS $= \dfrac{(x-1) + 2(x-2)}{(x-2)(x-1)} = \dfrac{3x-5}{x^2-3x+2}$. So $\dfrac{3x-5}{x^2-3x+2} = \dfrac{6}{x}$. Cross-multiplying: $x(3x-5) = 6(x^2-3x+2) \Rightarrow 3x^2 – 5x = 6x^2 – 18x + 12 \Rightarrow 3x^2 – 13x + 12 = 0$. Splitting $-13 = -9 – 4$ (product $36$): $(3x-4)(x-3) = 0$. So $x = \dfrac{4}{3}$ or $x = 3$.

Additional Practice — Long Answer / HOTS

Q1. The numerator of a fraction is 3 less than its denominator. If 1 is added to the denominator, the fraction is decreased by $\dfrac{1}{15}$. Find the fraction.

Answer: Let denominator $= x$, numerator $= (x-3)$. Original fraction $= \dfrac{x-3}{x}$. New fraction $= \dfrac{x-3}{x+1}$. Given $\dfrac{x-3}{x} – \dfrac{x-3}{x+1} = \dfrac{1}{15}$. LHS $= (x-3)\left[\dfrac{(x+1)-x}{x(x+1)}\right] = \dfrac{x-3}{x(x+1)}$. So $\dfrac{x-3}{x^2+x} = \dfrac{1}{15} \Rightarrow 15(x-3) = x^2 + x \Rightarrow x^2 – 14x + 45 = 0$. Splitting: $(x-5)(x-9)=0$. So $x = 5$ or $x = 9$. Both give valid fractions: $\dfrac{2}{5}$ or $\dfrac{6}{9} = \dfrac{2}{3}$.

Q2. A swimming pool is filled by three pipes with uniform flow. The first two pipes operating simultaneously fill the pool in the same time during which the pool is filled by the third pipe alone. The second pipe fills the pool 5 hours faster than the first pipe and 4 hours slower than the third pipe. Find the time required by each pipe.

Answer: Let the second pipe fill in $x$ hours. Then first $= (x+5)$ h, third $= (x-4)$ h. Given: $\dfrac{1}{x+5} + \dfrac{1}{x} = \dfrac{1}{x-4}$. Combine LHS: $\dfrac{x + (x+5)}{x(x+5)} = \dfrac{1}{x-4} \Rightarrow \dfrac{2x+5}{x^2+5x} = \dfrac{1}{x-4}$. Cross-multiplying: $(2x+5)(x-4) = x^2 + 5x \Rightarrow 2x^2 – 3x – 20 = x^2 + 5x \Rightarrow x^2 – 8x – 20 = 0$. Splitting: $(x-10)(x+2)=0$. Since $x > 4$, $x = 10$. Hence the times are 15 h, 10 h, and 6 h respectively.

Q3. A piece of cloth costs Rs 200. If the piece were 5 m longer and each metre cost Rs 2 less, the cost of the piece would have remained unchanged. How long is the piece and what is its original rate per metre?

Answer: Let length $= x$ m and rate $= y$ Rs/m. Then $xy = 200$ and $(x+5)(y-2) = 200$. From the first, $y = \dfrac{200}{x}$. Substituting: $(x+5)\left(\dfrac{200}{x} – 2\right) = 200 \Rightarrow (x+5) \cdot \dfrac{200 – 2x}{x} = 200 \Rightarrow (x+5)(200-2x) = 200x$. Expanding: $200x – 2x^2 + 1000 – 10x = 200x \Rightarrow -2x^2 – 10x + 1000 = 0 \Rightarrow x^2 + 5x – 500 = 0$. Splitting $5 = 25 – 20$: $(x+25)(x-20)=0$. Since $x > 0$, $x = 20$. Hence length = 20 m and rate = Rs 10/m.

Q4. Solve: $\sqrt{2x+9} + x = 13$.

Answer: Rearranging: $\sqrt{2x+9} = 13 – x$. Squaring: $2x + 9 = (13-x)^2 = 169 – 26x + x^2 \Rightarrow x^2 – 28x + 160 = 0$. Splitting $-28 = -8 – 20$ (product $160$): $(x-8)(x-20)=0$. So $x = 8$ or $x = 20$. Check: For $x=8$, LHS $=\sqrt{25}+8 = 13$ ✓. For $x=20$, LHS $=\sqrt{49}+20 = 27 \ne 13$. So only $x = 8$.

Q5. The difference of two natural numbers is 5 and the difference of their reciprocals is $\dfrac{1}{10}$. Find the numbers.

Answer: Let larger number be $x$ and smaller $(x-5)$. Given $\dfrac{1}{x-5} – \dfrac{1}{x} = \dfrac{1}{10} \Rightarrow \dfrac{x – (x-5)}{x(x-5)} = \dfrac{1}{10} \Rightarrow \dfrac{5}{x^2 – 5x} = \dfrac{1}{10} \Rightarrow x^2 – 5x = 50 \Rightarrow x^2 – 5x – 50 = 0$. Splitting: $(x-10)(x+5)=0$. Since $x \in \mathbb{N}$, $x = 10$. Hence numbers are 10 and 5.

Glossary / Key Terms

Term	Meaning
Quadratic equation	A polynomial equation of degree 2 in one variable; standard form $ax^2+bx+c=0,\ a\ne0$.
Standard form	$ax^2+bx+c=0$ with $a, b, c$ real and $a\ne 0$.
Root / Solution	A value of the variable that satisfies the equation.
Zero of polynomial	A value at which the polynomial equals zero — same as a root of the corresponding equation.
Factorisation	Splitting the quadratic into a product of two linear factors and equating each to zero.
Splitting the middle term	Writing $bx$ as the sum of two terms whose coefficients multiply to $ac$ and add to $b$.
Completing the square	Rewriting $ax^2+bx+c$ as $a\left(x+\dfrac{b}{2a}\right)^2 + \text{constant}$.
Quadratic formula	$x = \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$ — gives the roots directly.
Discriminant ($D$)	$D = b^2 – 4ac$ — controls the nature of the roots.
Nature of roots	$D>0$: distinct real; $D=0$: equal real; $D<0$: no real roots.
Sum of roots	$\alpha + \beta = -\dfrac{b}{a}$.
Product of roots	$\alpha\beta = \dfrac{c}{a}$.
Sridharacharya formula	Indian-origin name of the quadratic formula.
Coefficient	The numerical multiplier of a variable term ($a$ for $x^2$, $b$ for $x$).
Constant term	The term with no variable — denoted $c$ in $ax^2+bx+c=0$.

Worked Examples — Step-by-Step Walkthroughs

Example 1: Solve $6x^2 – x – 2 = 0$ by factorisation.

Solution: Here $a=6, b=-1, c=-2$. Product $ac = -12$. We need two numbers whose product is $-12$ and sum is $-1$: choose $-4$ and $3$. So $6x^2 – 4x + 3x – 2 = 0 \Rightarrow 2x(3x – 2) + 1(3x – 2) = 0 \Rightarrow (3x-2)(2x+1)=0$. Hence $x = \dfrac{2}{3}$ or $x = -\dfrac{1}{2}$.

Example 2: Find the roots of $3x^2 – 5x + 2 = 0$ by completing the square.

Solution: Divide by 3: $x^2 – \dfrac{5}{3}x + \dfrac{2}{3} = 0 \Rightarrow x^2 – \dfrac{5}{3}x = -\dfrac{2}{3}$. Add $\left(\dfrac{5}{6}\right)^2 = \dfrac{25}{36}$ to both sides: $\left(x – \dfrac{5}{6}\right)^2 = \dfrac{25}{36} – \dfrac{2}{3} = \dfrac{25 – 24}{36} = \dfrac{1}{36}$. Taking square root: $x – \dfrac{5}{6} = \pm \dfrac{1}{6}$. So $x = 1$ or $x = \dfrac{2}{3}$.

Example 3: Find the value of $k$ for which $kx^2 + 2x + 1 = 0$ has real and distinct roots.

Solution: For real distinct roots, $D > 0$. Here $a = k, b = 2, c = 1$. So $D = 4 – 4k > 0 \Rightarrow k < 1$. Also $k \ne 0$ (else not quadratic). Hence $k \in (-\infty, 0) \cup (0, 1)$.

Example 4: If one root of the quadratic equation $2x^2 + kx – 6 = 0$ is 2, find the value of $k$ and also find the other root.

Solution: Substituting $x = 2$: $2(4) + 2k – 6 = 0 \Rightarrow 8 + 2k – 6 = 0 \Rightarrow k = -1$. With $k = -1$, equation becomes $2x^2 – x – 6 = 0$. Splitting: $2x^2 – 4x + 3x – 6 = 0 \Rightarrow 2x(x-2) + 3(x-2) = 0 \Rightarrow (x-2)(2x+3) = 0$. Other root $= -\dfrac{3}{2}$.

Example 5: The sum of a number and its reciprocal is $\dfrac{17}{4}$. Find the number.

Solution: Let the number be $x$. Then $x + \dfrac{1}{x} = \dfrac{17}{4} \Rightarrow \dfrac{x^2+1}{x} = \dfrac{17}{4} \Rightarrow 4x^2 – 17x + 4 = 0$. Splitting $-17 = -16 – 1$ (product $16$): $4x^2 – 16x – x + 4 = 0 \Rightarrow 4x(x – 4) – 1(x-4) = 0 \Rightarrow (x-4)(4x-1) = 0$. So $x = 4$ or $x = \dfrac{1}{4}$.

Example 6: A two-digit number is such that the product of its digits is 18. When 63 is subtracted from the number, the digits interchange their places. Find the number.

Solution: Let the tens digit be $x$ and units digit $\dfrac{18}{x}$. Number $= 10x + \dfrac{18}{x}$. After subtracting 63: $10x + \dfrac{18}{x} – 63 = 10 \cdot \dfrac{18}{x} + x \Rightarrow 9x – \dfrac{162}{x} = 63$. Multiply by $x$: $9x^2 – 63x – 162 = 0 \Rightarrow x^2 – 7x – 18 = 0$. Splitting: $(x-9)(x+2)=0$. Since $x$ is a digit and positive, $x = 9$. So tens digit = 9, units digit = 2. Number is 92.

Example 7: The hypotenuse of a right triangle is $3\sqrt{10}$ cm. If the smaller side is tripled and the larger side is doubled, the new hypotenuse will be $9\sqrt{5}$ cm. Find the lengths of each side.

Solution: Let the smaller side $= x$ and larger side $= y$ cm. Then $x^2 + y^2 = 90$ and $(3x)^2 + (2y)^2 = 405 \Rightarrow 9x^2 + 4y^2 = 405$. From first: $y^2 = 90 – x^2$. Substituting: $9x^2 + 4(90-x^2) = 405 \Rightarrow 5x^2 = 45 \Rightarrow x^2 = 9 \Rightarrow x = 3$. Then $y^2 = 90 – 9 = 81 \Rightarrow y = 9$. Sides are 3 cm and 9 cm.

Fill in the Blanks

1. The standard form of a quadratic equation is __________.

Answer: $ax^2 + bx + c = 0$, where $a \ne 0$.

2. The discriminant of a quadratic equation is given by __________.

Answer: $D = b^2 – 4ac$.

3. If the discriminant is __________, the roots are real and equal.

Answer: zero (i.e. $D = 0$).

4. The roots of $x^2 – 4 = 0$ are __________.

Answer: $x = \pm 2$.

5. The sum of the roots of $ax^2 + bx + c = 0$ is __________.

Answer: $-\dfrac{b}{a}$.

6. A quadratic equation has at most __________ roots.

Answer: two.

7. If $D < 0$, the quadratic equation has __________ roots.

Answer: no real.

8. The product of the roots of $5x^2 – 6x + 2 = 0$ is __________.

Answer: $\dfrac{2}{5}$.

True or False

1. Every quadratic equation has at least one real root.

Answer: False — when $D < 0$, no real roots exist.

2. A quadratic equation can have three real roots.

Answer: False — at most two roots.

3. The roots of $x^2 + 1 = 0$ are real.

Answer: False — $D = -4 < 0$, no real roots.

4. The equation $3x^2 – 6x + 3 = 0$ has equal roots.

Answer: True — $D = 36 – 36 = 0$.

5. If $a + b + c = 0$, then $x = 1$ is always a root of $ax^2 + bx + c = 0$.

Answer: True — substituting $x = 1$ gives $a + b + c$ which is $0$.

6. The quadratic equation $x^2 + x + 1 = 0$ has two distinct real roots.

Answer: False — $D = 1 – 4 = -3 < 0$.

Common Mistakes & Tips

Always reduce to standard form first. Move every term to one side so the equation looks like $ax^2 + bx + c = 0$ before applying any method.
Check the discriminant before solving. If $D < 0$, do not waste time on the quadratic formula — declare “no real roots”.
Verify which root is admissible. Word problems frequently produce two algebraic solutions, but a negative age, a negative speed, or a non-integer count of toys must be rejected.
For “$D = 0$” or “equal roots” type questions, set the discriminant to zero and solve for the parameter — this is a recurring board-exam pattern.
Beware the $a \ne 0$ trap. If a parameter $k$ multiplies $x^2$ and you find $k = 0$ as one solution, reject it unless the equation is allowed to be linear.
Splitting the middle term needs $ac$, not just $c$. When the leading coefficient is not 1, multiply $a \times c$ to find the product whose factors must add to $b$.
Always cross-check by substituting your final root back into the original equation — this catches sign errors and arithmetic slips.
Squaring both sides may introduce extraneous roots. When solving radical equations like $\sqrt{2x+9} + x = 13$, always verify each candidate.

Quick Revision Box

Standard form: $ax^2 + bx + c = 0,\ a \ne 0$.
Quadratic formula: $x = \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$.
Discriminant: $D = b^2 – 4ac$.
$D > 0$ → two distinct real roots.
$D = 0$ → two equal real roots, $x = -\dfrac{b}{2a}$.
$D < 0$ → no real roots.
Sum of roots: $\alpha + \beta = -\dfrac{b}{a}$.
Product of roots: $\alpha \beta = \dfrac{c}{a}$.
Equation with given roots: $x^2 – (\alpha+\beta)x + \alpha\beta = 0$.
Three solving methods — factorisation, completing the square, quadratic formula — give the same roots.

Examination-Pattern Mixed Practice

1-Mark Type:

Write the discriminant of $3x^2 – 2x + 8 = 0$. Answer: $D = 4 – 96 = -92$.
Find the nature of roots of $2x^2 + 5\sqrt{3}\,x + 6 = 0$. Answer: $D = 75 – 48 = 27 > 0$, two distinct real roots.
If $x = -\dfrac{1}{2}$ is a root of $2x^2 + bx – 2 = 0$, find $b$. Answer: $\dfrac{1}{2} – \dfrac{b}{2} – 2 = 0 \Rightarrow b = -3$.
Roots of $9x^2 – 3x – 2 = 0$? Answer: $(3x-2)(3x+1)/3 = 0$ gives $x = \dfrac{2}{3}, -\dfrac{1}{3}$.
Sum of roots of $x^2 – 3x + 2 = 0$? Answer: $3$.

2-Mark Type:

Solve $x^2 + 5x – 6 = 0$ by factorisation. Answer: $(x+6)(x-1)=0 \Rightarrow x = -6, 1$.
Find $k$ if $x = 3$ is a root of $kx^2 – 5x – 3 = 0$. Answer: $9k – 15 – 3 = 0 \Rightarrow k = 2$.
For what value of $p$ does $px^2 – 4x + 3 = 0$ have equal roots? Answer: $D = 16 – 12p = 0 \Rightarrow p = \dfrac{4}{3}$.

3-Mark Type:

Solve $4x^2 – 4ax + (a^2 – b^2) = 0$ for $x$. Answer: $D = 16a^2 – 16(a^2-b^2) = 16b^2$. $x = \dfrac{4a \pm 4b}{8} = \dfrac{a \pm b}{2}$.
Solve $\dfrac{x-1}{x-2} + \dfrac{x-3}{x-4} = \dfrac{10}{3}$, $x \ne 2, 4$. Answer: Combine LHS: $\dfrac{(x-1)(x-4) + (x-2)(x-3)}{(x-2)(x-4)} = \dfrac{10}{3}$. Numerator $= 2x^2 – 10x + 10$, denominator $= x^2 – 6x + 8$. So $3(2x^2 – 10x + 10) = 10(x^2 – 6x + 8) \Rightarrow 6x^2 – 30x + 30 = 10x^2 – 60x + 80 \Rightarrow 4x^2 – 30x + 50 = 0 \Rightarrow 2x^2 – 15x + 25 = 0$. Splitting: $(2x-5)(x-5) = 0$. $x = \dfrac{5}{2}$ or $x = 5$.

4-Mark Type:

The difference between two numbers is 9 and the difference of their squares is 207. Find the numbers. Answer: Let the larger be $x$, smaller $(x-9)$. Then $x^2 – (x-9)^2 = 207 \Rightarrow 18x – 81 = 207 \Rightarrow x = 16$. Numbers are 16 and 7.
Two pipes running together can fill a tank in $11\dfrac{1}{9}$ minutes. If one pipe takes 5 minutes more than the other, find the times. Answer: Total time $= \dfrac{100}{9}$ min. Let smaller take $x$ min: $\dfrac{1}{x} + \dfrac{1}{x+5} = \dfrac{9}{100}$. After simplification: $9x^2 – 155x – 500 = 0 \Rightarrow x = 20$ (positive). Pipes take 20 min and 25 min.

Historical Note — The Quadratic Through the Ages

Quadratic equations have fascinated mathematicians for over four millennia. Babylonian clay tablets dating to around 2000 BCE record solutions to specific quadratics using a procedure equivalent to completing the square. The Greek mathematician Euclid (around 300 BCE) gave geometric solutions in his Elements, treating quadratics as area problems on rectangles and squares.

In India, the great mathematician Brahmagupta (598–668 CE) gave the first explicit symbolic formula for solving quadratic equations in his Brahma-sphuta-siddhanta. He stated rules for handling positive and negative quantities and accepted negative solutions — a major step beyond the Greeks. Sridharacharya (around 1025 CE) is traditionally credited with the modern compact form of the quadratic formula, $x = \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$, which is why it is sometimes called Sridhara’s formula in Indian textbooks. Bhaskara II (1114–1185 CE), in his celebrated Bijaganita, refined these ideas and gave systematic word problems much like the ones you have just solved.

The Persian mathematician al-Khwarizmi (around 820 CE) wrote Al-Kitāb al-mukhtaṣar fī ḥisāb al-jabr wal-muqābala, from which the very word algebra derives. He classified quadratics into six standard cases (depending on signs of coefficients) and explained each by completing the square.

So when you solve a quadratic equation today, you are using a tool sharpened over four thousand years across Babylon, Greece, India, the Islamic world, and modern Europe — a beautiful example of how mathematical knowledge is genuinely a shared inheritance of humankind.

Application to Real-Life Situations

Quadratic equations are not abstract symbol-shuffling — they describe an enormous range of phenomena where one quantity depends on the square of another. Let us look at a few classic application areas.

Projectile motion (Physics): When a ball is thrown straight up with initial speed $u$ from a height $h_0$, its height after time $t$ seconds is $h(t) = h_0 + ut – \dfrac{1}{2}gt^2$, where $g \approx 9.8$ m/s² is gravity. Setting $h(t) = 0$ gives a quadratic in $t$ — the time of flight.

Area problems (Geometry): If you double the side of a square garden and the new area exceeds the old by $300\,\text{m}^2$, you instantly get a quadratic in the original side: $(2x)^2 – x^2 = 300 \Rightarrow 3x^2 = 300 \Rightarrow x = 10$. (This particular case is “linear-in-square”, but most area problems are genuine quadratics.)

Speed–Time–Distance: If a vehicle’s average speed and a slight change in it produce a difference in time, the equation invariably becomes quadratic in the speed (since time $=$ distance ÷ speed introduces a reciprocal).

Profit–Cost analysis (Economics): If profit per unit varies linearly with the number of units sold, total profit is a quadratic in the number of units, and the maximum-profit point is found by setting the derivative to zero — equivalent to using the formula $x = -\dfrac{b}{2a}$ for the vertex of a parabola.

Number puzzles: “The square of a number plus twice the number is 24” is just $x^2 + 2x – 24 = 0$, factoring to $(x+6)(x-4)=0$.

Final Solved Example — Putting It All Together

Problem: A bus travels 300 km at a constant speed. If its speed had been 5 km/h more, it would have taken 2 hours less for the journey. Find the original speed of the bus.

Solution: Let the original speed be $x$ km/h. Then time $= \dfrac{300}{x}$ h. With increased speed: time $= \dfrac{300}{x+5}$ h. Given $\dfrac{300}{x} – \dfrac{300}{x+5} = 2$. Multiply by $x(x+5)$: $300(x+5) – 300x = 2x(x+5) \Rightarrow 1500 = 2x^2 + 10x \Rightarrow x^2 + 5x – 750 = 0$.

Apply the quadratic formula: $a = 1, b = 5, c = -750$. $D = 25 + 3000 = 3025$, $\sqrt{D} = 55$. $x = \dfrac{-5 \pm 55}{2}$. Positive root: $x = \dfrac{50}{2} = 25$. Hence the original speed of the bus is 25 km/h.

Verification: At 25 km/h, time $= \dfrac{300}{25} = 12$ h. At 30 km/h, time $= \dfrac{300}{30} = 10$ h. Difference $= 2$ h. ✓

This completes the comprehensive English-medium guide to ASSEB Class 10 Mathematics Chapter 4 — Quadratic Equations. Practise each problem carefully, memorise the quadratic formula, and master the discriminant test for the nature of roots. Wishing you the very best for your HSLC examination from HSLC GURU!