NCERT Exemplar Solutions Class 10 Maths Quadratic Equations — Exercises 4.3 and 4.4

NCERT Exemplar Problems, Class 10 Mathematics — Quadratic Equations, Exercises 4.3 and 4.4. All 10 questions solved.

Exercise 4.3 is two questions with twelve equations between them — seven to be solved by the quadratic formula and five by factorisation — and Exercise 4.4 is eight questions, five more equations and then the word problems. They are taken together because Exercise 4.3 has only two numbered questions.

The formula is

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

and two habits make it reliable. Clear the fractions and surds from the coefficients first — multiplying the whole equation by a constant leaves the roots untouched and makes the arithmetic integer. And simplify the surd in $\sqrt{D}$ before dividing: $\sqrt{405} = 9\sqrt5$, and the $\sqrt5$ then cancels against the numerator’s other term.

In the word problems of Exercise 4.4, every quadratic has two roots and usually only one of them is an answer. Rejecting the other is part of the solution, not an afterthought — a length cannot be negative, an age cannot be a fraction of a year here, and a border cannot be wider than the lawn.

Key insight. A quadratic gives two candidate answers. In a word problem, go back to the sentence the equation came from and check which candidate the sentence actually permits; in Question 5 of Exercise 4.4 both roots are positive whole numbers, and only the wording rules one out.

Exercise 4.3, Question 1 — Find the roots of the quadratic equations by using the quadratic formula in each of the following.

Exercise 4.3, Question 1

(i) $2x^2 – 3x – 5 = 0$

$D = 9 + 40 = 49$, $\sqrt D = 7$.

$$x = \frac{3 \pm 7}{4} = \frac{10}{4} \text{ or } \frac{-4}{4} = \frac52 \text{ or } -1$$

(ii) $5x^2 + 13x + 8 = 0$

$D = 169 – 160 = 9$, $\sqrt D = 3$.

$$x = \frac{-13 \pm 3}{10} = -1 \text{ or } -\frac{8}{5}$$

(iii) $-3x^2 + 5x + 12 = 0$

Multiply by $-1$ so the leading coefficient is positive: $3x^2 – 5x – 12 = 0$. Then $D = 25 + 144 = 169$, $\sqrt D = 13$.

$$x = \frac{5 \pm 13}{6} = 3 \text{ or } -\frac43$$

(iv) $-x^2 + 7x – 10 = 0$

Multiply by $-1$: $x^2 – 7x + 10 = 0$. $D = 49 – 40 = 9$.

$$x = \frac{7 \pm 3}{2} = 5 \text{ or } 2$$

(v) $x^2 + 2\sqrt2 x – 6 = 0$

$D = \left(2\sqrt2\right)^2 + 24 = 8 + 24 = 32$, and $\sqrt{32} = 4\sqrt2$.

$$x = \frac{-2\sqrt2 \pm 4\sqrt2}{2} = \frac{2\sqrt2}{2} \text{ or } \frac{-6\sqrt2}{2} = \sqrt2 \text{ or } -3\sqrt2$$

(vi) $x^2 – 3\sqrt5 x + 10 = 0$

$D = 45 – 40 = 5$, $\sqrt D = \sqrt5$.

$$x = \frac{3\sqrt5 \pm \sqrt5}{2} = \frac{4\sqrt5}{2} \text{ or } \frac{2\sqrt5}{2} = 2\sqrt5 \text{ or } \sqrt5$$

(vii) $\dfrac12 x^2 – \sqrt{11}\,x + 1 = 0$

Multiply by $2$ first: $x^2 – 2\sqrt{11}x + 2 = 0$. Then $D = 44 – 8 = 36$, $\sqrt D = 6$.

$$x = \frac{2\sqrt{11} \pm 6}{2} = \sqrt{11} + 3 \text{ or } \sqrt{11} – 3$$

Exercise 4.3, Question 2

Find the roots of the following quadratic equations by the factorisation method.

(i) $2x^2 + \dfrac53 x – 2 = 0$

Multiply by $3$: $6x^2 + 5x – 6 = 0$. Here $ac = -36$ and $b = 5$, so split with $9$ and $-4$:

$$6x^2 + 9x – 4x – 6 = 3x(2x+3) – 2(2x+3) = (2x+3)(3x-2)$$

Roots: $-\dfrac32$ and $\dfrac23$.

(ii) $\dfrac25 x^2 – x – \dfrac35 = 0$

Multiply by $5$: $2x^2 – 5x – 3 = 0$. $ac = -6$, $b = -5$, so split with $-6$ and $1$:

$$2x^2 – 6x + x – 3 = 2x(x-3) + 1(x-3) = (x-3)(2x+1)$$

Roots: $3$ and $-\dfrac12$.

(iii) $3\sqrt2 x^2 – 5x – \sqrt2 = 0$

$ac = 3\sqrt2 \times \left(-\sqrt2\right) = -6$ — rational, which is what makes the search possible. Two numbers with product $-6$ and sum $-5$: $-6$ and $1$.

$$3\sqrt2 x^2 – 6x + x – \sqrt2 = 3\sqrt2 x\left(x – \sqrt2\right) + 1\left(x – \sqrt2\right) = \left(x – \sqrt2\right)\left(3\sqrt2 x + 1\right)$$

using $3\sqrt2 x \times \left(-\sqrt2\right) = -6x$. The roots are $x = \sqrt2$ and

$$x = -\frac{1}{3\sqrt2} = -\frac{\sqrt2}{6}$$

(iv) $3x^2 + 5\sqrt5 x – 10 = 0$

$ac = -30$ and $b = 5\sqrt5$. Two surds with product $-30$ and sum $5\sqrt5$: $6\sqrt5$ and $-\sqrt5$, since $6\sqrt5 \times \left(-\sqrt5\right) = -30$.

$$3x^2 + 6\sqrt5 x – \sqrt5 x – 10 = 3x\left(x + 2\sqrt5\right) – \sqrt5\left(x + 2\sqrt5\right) = \left(x + 2\sqrt5\right)\left(3x – \sqrt5\right)$$

Roots: $-2\sqrt5$ and $\dfrac{\sqrt5}{3}$.

(v) $21x^2 – 2x + \dfrac{1}{21} = 0$

Multiply by $21$: $441x^2 – 42x + 1 = 0$. This is a perfect square:

$$(21x)^2 – 2(21x)(1) + 1^2 = (21x – 1)^2 = 0$$

So there is a repeated root $x = \dfrac{1}{21}$, twice. (Consistently, $D = 42^2 – 4(441) = 1764 – 1764 = 0$.)

Exercise 4.4

Exercise 4.4, Question 1

Find whether the following equations have real roots. If real roots exist, find them.

(i) $8x^2 + 2x – 3 = 0$

$D = 4 + 96 = 100 > 0$ — real and distinct. $\sqrt D = 10$:

$$x = \frac{-2 \pm 10}{16} = \frac12 \text{ or } -\frac34$$

(ii) $-2x^2 + 3x + 2 = 0$

Multiply by $-1$: $2x^2 – 3x – 2 = 0$. $D = 9 + 16 = 25 > 0$:

$$x = \frac{3 \pm 5}{4} = 2 \text{ or } -\frac12$$

(iii) $5x^2 – 2x – 10 = 0$

$D = 4 + 200 = 204 > 0$. Now $\sqrt{204} = \sqrt{4 \times 51} = 2\sqrt{51}$, and the factor of $2$ is what makes the answer tidy:

$$x = \frac{2 \pm 2\sqrt{51}}{10} = \frac{1 \pm \sqrt{51}}{5}$$

(iv) $\dfrac{1}{2x-3} + \dfrac{1}{x-5} = 1$, where $x \ne \dfrac32$, $5$

Multiply through by $(2x-3)(x-5)$:

$$(x – 5) + (2x – 3) = (2x-3)(x-5)$$ $$3x – 8 = 2x^2 – 13x + 15$$ $$2x^2 – 16x + 23 = 0$$

$D = 256 – 184 = 72 > 0$, and $\sqrt{72} = 6\sqrt2$:

$$x = \frac{16 \pm 6\sqrt2}{4} = 4 \pm \frac{3\sqrt2}{2}$$

Neither value is $\frac32$ or $5$, so both are admissible.

(v) $x^2 + 5\sqrt5 x – 70 = 0$

$D = 125 + 280 = 405 > 0$, and $\sqrt{405} = \sqrt{81 \times 5} = 9\sqrt5$:

$$x = \frac{-5\sqrt5 \pm 9\sqrt5}{2} = \frac{4\sqrt5}{2} \text{ or } \frac{-14\sqrt5}{2} = 2\sqrt5 \text{ or } -7\sqrt5$$

Exercise 4.4, Question 2

Find a natural number whose square diminished by $84$ is equal to thrice of $8$ more than the given number.

Solution. Let the number be $n$. “Square diminished by $84$” is $n^2 – 84$; “thrice of $8$ more than the number” is $3(n + 8)$.

$$n^2 – 84 = 3(n + 8) = 3n + 24$$ $$n^2 – 3n – 108 = 0$$

Factorising ($-12$ and $9$ have product $-108$ and sum $-3$):

$$(n – 12)(n + 9) = 0 \implies n = 12 \text{ or } n = -9$$

A natural number cannot be negative, so $n = 12$. Check: $144 – 84 = 60$ and $3(12 + 8) = 60$ ✓.

Exercise 4.4, Question 3

A natural number, when increased by $12$, equals $160$ times its reciprocal. Find the number.

Solution. Let the number be $n$; its reciprocal is $\frac1n$.

$$n + 12 = \frac{160}{n}$$

Multiplying by $n$ (legitimate since $n \ne 0$):

$$n^2 + 12n – 160 = 0 \implies (n + 20)(n – 8) = 0$$

so $n = 8$ or $n = -20$, and only $8$ is natural. Check: $8 + 12 = 20$ and $\frac{160}{8} = 20$ ✓.

Exercise 4.4, Question 4

A train, travelling at a uniform speed for $360$ km, would have taken $48$ minutes less to travel the same distance if its speed were $5$ km/h more. Find the original speed of the train.

Solution. Let the original speed be $s$ km/h. The two journey times are $\frac{360}{s}$ and $\frac{360}{s+5}$ hours, and the faster one is shorter by $48$ minutes $= \frac{48}{60} = \frac45$ hour.

$$\frac{360}{s} – \frac{360}{s+5} = \frac45$$

Combine the left side over a common denominator:

$$\frac{360(s+5) – 360s}{s(s+5)} = \frac{1800}{s(s+5)} = \frac45$$

Cross-multiplying, $9000 = 4s(s+5)$, that is

$$4s^2 + 20s – 9000 = 0 \implies s^2 + 5s – 2250 = 0 \implies (s + 50)(s – 45) = 0$$

A speed cannot be negative, so $s = 45$.

Check: $\frac{360}{45} = 8$ hours and $\frac{360}{50} = 7.2$ hours, a difference of $0.8$ hour $= 48$ minutes ✓.

Exercise 4.4, Question 5

If Zeba were younger by $5$ years than what she really is, then the square of her age (in years) would have been $11$ more than five times her actual age. What is her age now?

Solution. Let her actual age be $z$ years. Five years younger she would be $z – 5$, and the square of that age is what the sentence talks about:

$$(z – 5)^2 = 5z + 11$$ $$z^2 – 10z + 25 = 5z + 11$$ $$z^2 – 15z + 14 = 0 \implies (z – 1)(z – 14) = 0$$

so $z = 14$ or $z = 1$. Both are positive whole numbers, so the arithmetic does not decide between them — the wording does. “Younger by $5$ years than what she really is” only makes sense if she is older than $5$, and $z = 1$ would put her age five years ago at $-4$. Hence $z = 14$.

Check: $(14-5)^2 = 81$ and $5(14) + 11 = 81$ ✓.

Exercise 4.4, Question 6

At present Asha’s age (in years) is $2$ more than the square of her daughter Nisha’s age. When Nisha grows to her mother’s present age, Asha’s age would be one year less than $10$ times the present age of Nisha. Find the present ages of both Asha and Nisha.

Solution. Let Nisha be $n$ years old now, so Asha is $n^2 + 2$.

The second sentence needs care. Nisha reaches her mother’s present age after

$$\left(n^2 + 2\right) – n \text{ years}$$

and Asha ages by the same amount in that time, so Asha’s age then is

$$\left(n^2 + 2\right) + \left[\left(n^2+2\right) – n\right] = 2\left(n^2+2\right) – n$$

Setting this equal to $10n – 1$:

$$2n^2 + 4 – n = 10n – 1 \implies 2n^2 – 11n + 5 = 0$$

Factorising ($ac = 10$, split $-11$ as $-10$ and $-1$):

$$2n^2 – 10n – n + 5 = 2n(n-5) – 1(n-5) = (n-5)(2n-1) = 0$$

so $n = 5$ or $n = \frac12$. An age of half a year makes the mother $2\frac14$ years old, which is impossible, so $n = 5$ and Asha is $5^2 + 2 = 27$.

Check: Nisha reaches $27$ in $22$ years, when Asha is $49$; and $10(5) – 1 = 49$ ✓.

Exercise 4.4, Question 7

In the centre of a rectangular lawn of dimensions $50$ m $\times$ $40$ m, a rectangular pond has to be constructed so that the area of the grass surrounding the pond would be $1184$ m². Find the length and breadth of the pond.

Solution. The pond is in the centre, so the grass border has the same width all the way round — call it $w$ metres. That single letter is what makes the problem a quadratic rather than a pair of equations.

The pond then measures $(50 – 2w)$ by $(40 – 2w)$, losing $w$ at each of two opposite edges. Its area is the lawn’s area minus the grass:

$$\text{pond area} = 50 \times 40 – 1184 = 2000 – 1184 = 816$$

So

$$(50 – 2w)(40 – 2w) = 816$$ $$2000 – 100w – 80w + 4w^2 = 816$$ $$4w^2 – 180w + 1184 = 0 \implies w^2 – 45w + 296 = 0$$

$D = 2025 – 1184 = 841 = 29^2$, so

$$w = \frac{45 \pm 29}{2} = 37 \text{ or } 8$$

A border $37$ m wide is impossible — twice that already exceeds both sides of the lawn — so $w = 8$, and the pond is

$$50 – 16 = 34 \text{ m by } 40 – 16 = 24 \text{ m}$$

Check: $34 \times 24 = 816$ ✓, so the grass is $2000 – 816 = 1184$ m² ✓.

Exercise 4.4, Question 8

At $t$ minutes past $2$ pm, the time needed by the minute hand of a clock to show $3$ pm was found to be $3$ minutes less than $\dfrac{t^2}{4}$ minutes. Find $t$.

Solution. From $2$ pm to $3$ pm is $60$ minutes. At $t$ minutes past $2$, the time still needed is $60 – t$ minutes. The sentence says this equals $\frac{t^2}{4}$ less $3$:

$$60 – t = \frac{t^2}{4} – 3$$

Multiply by $4$:

$$240 – 4t = t^2 – 12 \implies t^2 + 4t – 252 = 0$$

$D = 16 + 1008 = 1024 = 32^2$, so

$$t = \frac{-4 \pm 32}{2} = 14 \text{ or } -18$$

A time past $2$ pm cannot be negative, so $t = 14$.

Check: at $2{:}14$ there are $46$ minutes to go, and $\frac{14^2}{4} – 3 = 49 – 3 = 46$ ✓.

Common mistakes

• Leaving fractions or surds in the coefficients. Exercise 4.3 Question 1(vii) and Question 2(i), (ii), (v) all become integer equations after one multiplication, and the formula is far less error-prone that way.
• Not simplifying $\sqrt D$ before dividing. $\sqrt{204} = 2\sqrt{51}$ and $\sqrt{405} = 9\sqrt5$; without pulling the square factor out, the fraction will not cancel and the answer looks nothing like the printed one.
• Missing that a root is repeated in Exercise 4.3 Question 2(v). $441x^2 – 42x + 1$ is $(21x-1)^2$, so $\frac{1}{21}$ is a double root. Writing a single root is not quite the same answer.
• Keeping the inadmissible root. Every word problem here produces two, and one must be argued away: negative numbers in Questions 2, 3, 4 and 8, a fractional age in Question 6, an impossible border width in Question 7. Say why it is rejected — the reason carries marks.
• Rejecting on the wrong grounds in Question 5. Both roots, $1$ and $14$, are positive whole numbers, so “age must be positive” does not settle it. What settles it is that “five years younger” presupposes she is at least five.
• Using two variables for the pond in Question 7. The word “centre” forces a uniform border, so the pond is $(50-2w)$ by $(40-2w)$ with one unknown. Treating length and breadth as independent gives one equation and two unknowns.
• Subtracting $w$ once instead of twice. The border eats into both ends of each side, so $50$ becomes $50 – 2w$.
• Forgetting to exclude the forbidden values in Exercise 4.4 Question 1(iv). The equation is undefined at $x = \frac32$ and $x = 5$; both roots must be checked against them, even though neither turns out to be excluded here.

Practise next

• Exemplar Exercises 4.1 and 4.2 — the discriminant test on its own, which is what Question 1 of Exercise 4.4 applies before solving.
• NCERT textbook Exercise 4.2 — factorisation on the book’s own equations, including several word problems of the same shape as Questions 2 to 6 above.
• NCERT textbook Exercise 4.3 for the nature of the roots and the formula, and Exercise 4.1 for forming a quadratic from a situation — the step Questions 4, 7 and 8 above turn on.
• Exemplar Exercise 3.4 for the same speed–distance–time setups when they stay linear.