NCERT Exemplar Solutions Class 10 Maths Quadratic Equations — Exercises 4.1 and 4.2

NCERT Exemplar Problems, Class 10 Mathematics — Quadratic Equations, Exercises 4.1 and 4.2. All 18 questions solved.

Exercise 4.1 is eleven multiple-choice questions and Exercise 4.2 is seven reasoning questions with twenty-one sub-parts, and the two together rest on two ideas.

First, what counts as a quadratic equation. An equation is quadratic only after it has been simplified to the form $ax^2 + bx + c = 0$ with $a \ne 0$. Several questions here present equations that look quadratic and are not, because the $x^2$ terms cancel — and one that looks quartic and is not.

Second, the discriminant $D = b^2 – 4ac$, which answers every question about the nature of the roots without finding them:

$D$	Roots
$D > 0$	two distinct real roots
$D = 0$	two equal real roots
$D < 0$	no real roots

A note on the current syllabus. Question 7 of Exercise 4.1 is about completing the square, and that method is no longer in the Class 10 textbook: the chapter runs Introduction → Quadratic Equations → Solution by Factorisation → Nature of Roots → Summary, and the phrase appears just once, in the historical note about Sridharacharya. The quadratic formula itself is retained and used. Everything else on this page is squarely on syllabus.

Key insight. Simplify before you classify. More than half the errors on this page come from reading the printed form of an equation rather than the form it reduces to — and the exemplar has laid the traps deliberately.

Choose the correct answer from the given four options:

Exercise 4.1, Question 1

Which of the following is a quadratic equation?

Solution. Expand each and see what survives.

• (A) $x^2 + 2x + 1 = 16 – 8x + x^2 + 3$. The $x^2$ cancels, leaving $10x – 18 = 0$ — linear.
• (B) The right side is $10x – 2 – 2x^2 + \frac25 x$, so the $-2x^2$ cancels against the left, leaving $2 – \frac{52}{5}x = 0$ — linear.
• (C) With $k = -1$ the coefficient $k + 1$ is zero, so the equation is $\frac32 x = 7$ — linear. The “where $k = -1$” is the whole question.
• (D) $x^3 – x^2 = x^3 – 3x^2 + 3x – 1$. The $x^3$ cancels, leaving $2x^2 – 3x + 1 = 0$ — quadratic ✓

Exercise 4.1, Question 2

Which of the following is not a quadratic equation?

Solution. Again, expand.

• (A) $2x^2 – 4x + 2 = 4x^2 – 2x + 1$, that is $2x^2 + 2x – 1 = 0$ — quadratic.
• (B) $2x – x^2 = x^2 + 5$, that is $2x^2 – 2x + 5 = 0$ — quadratic.
• (C) $\left(\sqrt2x+\sqrt3\right)^2 = 2x^2 + 2\sqrt6 x + 3$, so the left side is $3x^2 + 2\sqrt6x + 3$. Setting this equal to $3x^2 – 5x$, the $3x^2$ cancels and what remains is $2\sqrt6 x + 5x + 3 = 0$ — linear ✓
• (D) $x^4 + 4x^3 + 4x^2 = x^4 + 4x^3 + 3$, so $4x^2 – 3 = 0$ — quadratic, despite both sides being quartic.

Exercise 4.1, Question 3

Which of the following equations has $2$ as a root?

Solution. A root makes the left side zero, so substitute $x = 2$ in each — four quick arithmetic checks, and no solving at all.

$$\text{(A) } 4 – 8 + 5 = 1, \qquad \text{(B) } 4 + 6 – 12 = -2$$ $$\text{(C) } 8 – 14 + 6 = 0 \ ✓, \qquad \text{(D) } 12 – 12 – 2 = -2$$

Exercise 4.1, Question 4

If $\dfrac12$ is a root of the equation $x^2 + kx – \dfrac54 = 0$, then the value of $k$ is

Solution. Substitute $x = \frac12$ and solve for $k$:

$$\frac14 + \frac{k}{2} – \frac54 = 0 \implies \frac{k}{2} = 1 \implies k = 2$$

Exercise 4.1, Question 5

Which of the following equations has the sum of its roots as $3$?

Solution. The sum of the roots is $-\dfrac{b}{a}$, so this is four divisions.

$$\text{(A) } \frac{3}{2}, \qquad \text{(B) } \frac{-3}{-1} = 3 \ ✓, \qquad \text{(C) } \frac{3/\sqrt2}{\sqrt2} = \frac{3}{2}, \qquad \text{(D) } \frac{3}{3} = 1$$

Note that the sum of the roots is defined whether or not the roots are real. Option (B) has discriminant $9 – 12 = -3 < 0$, so its “roots” are not real numbers — but the sum of the roots is still $3$, and the question asks only for that.

Exercise 4.1, Question 6

Values of $k$ for which the quadratic equation $2x^2 – kx + k = 0$ has equal roots is

Solution. Equal roots means $D = 0$:

$$(-k)^2 – 4(2)(k) = 0 \implies k^2 – 8k = 0 \implies k(k – 8) = 0$$

so $k = 0$ or $k = 8$. Both are legitimate: at $k = 0$ the equation is $2x^2 = 0$, whose roots are $0$ and $0$ — equal, and the equation is still quadratic because the coefficient of $x^2$ is $2$, not $k$.

Options (A) and (C) each keep one root of $k(k-8) = 0$ and discard the other; the word “only” in both is what makes them wrong.

Exercise 4.1, Question 7

Which constant must be added and subtracted to solve the quadratic equation $9x^2 + \dfrac34 x – \sqrt2 = 0$ by the method of completing the square?

Solution. Completing the square means matching the first two terms to $(\text{something})^2 = A^2 + 2Ac + c^2$, and then supplying the missing $c^2$.

Here $9x^2 = (3x)^2$, so $A = 3x$. Matching the linear term:

$$2Ac = 2(3x)c = 6cx = \frac34 x \implies c = \frac18$$

The constant to be added and subtracted is $c^2$:

$$c^2 = \left(\frac18\right)^2 = \frac{1}{64}$$

Option (A) is $c$ itself rather than $c^2$ — the commonest slip, and the reason both numbers appear among the options.

Syllabus note: as explained at the top of this page, completing the square is no longer taught in the Class 10 textbook, which solves quadratics by factorisation and by the formula. The technique is still worth seeing, because it is where the quadratic formula comes from.

Exercise 4.1, Question 8

The quadratic equation $2x^2 – \sqrt5 x + 1 = 0$ has

Solution.

$$D = \left(-\sqrt5\right)^2 – 4(2)(1) = 5 – 8 = -3 < 0$$

so there are no real roots. Note that $\left(-\sqrt5\right)^2 = 5$, not $\sqrt5$ — squaring removes the surd entirely, which is what makes these discriminants tidy.

Option (D) is impossible for any quadratic: it can have at most two roots.

Exercise 4.1, Question 9

Which of the following equations has two distinct real roots?

Solution. Four discriminants:

• (A) $\left(3\sqrt2\right)^2 – 4(2)\left(\frac94\right) = 18 – 18 = 0$ — equal roots, not distinct.
• (B) $1 + 20 = 21 > 0$ ✓
• (C) $9 – 8\sqrt2 \approx 9 – 11.31 = -2.31 < 0$ — no real roots. This one needs care: $8\sqrt2$ is a little over $11$, so the sign is not obvious by eye.
• (D) $9 – 20 = -11 < 0$.

Exercise 4.1, Question 10

Which of the following equations has no real roots?

Solution. Options (B) and (C) have a negative constant term, so $-4ac > 0$ and the discriminant is certainly positive — both can be discarded without arithmetic. That leaves (A) and (D):

• (A) $16 – 12\sqrt2 \approx 16 – 16.97 = -0.97 < 0$ ✓ — no real roots, but only just, so the estimate $\sqrt2 \approx 1.414$ has to be used carefully: $12 \times 1.414 = 16.97 > 16$.
• (D) $\left(4\sqrt3\right)^2 – 4(3)(4) = 48 – 48 = 0$ — two equal real roots, which are still real.

Exercise 4.1, Question 11

$\left(x^2 + 1\right)^2 – x^2 = 0$ has

Solution. Expand first:

$$\left(x^2+1\right)^2 – x^2 = x^4 + 2x^2 + 1 – x^2 = x^4 + x^2 + 1$$

This is a quartic, but it is quadratic in $x^2$. Putting $u = x^2$:

$$u^2 + u + 1 = 0, \qquad D = 1 – 4 = -3 < 0$$

so there is no real $u$ at all, and hence no real $x$.

A one-line alternative: for every real $x$, $x^4 \ge 0$ and $x^2 \ge 0$, so $x^4 + x^2 + 1 \ge 1 > 0$. The expression is never zero.

Exercise 4.2, Question 1

State whether the following quadratic equations have two distinct real roots. Justify your answer.

The test is $D = b^2 – 4ac > 0$ in every case, so this is one calculation repeated ten times.

(i) $x^2 – 3x + 4 = 0$

$D = 9 – 16 = -7 < 0$ — No, no real roots at all.

(ii) $2x^2 + x – 1 = 0$

$D = 1 + 8 = 9 > 0$ — Yes, two distinct real roots.

(iii) $2x^2 – 6x + \dfrac92 = 0$

$D = 36 – 4(2)\left(\frac92\right) = 36 – 36 = 0$ — No. The roots are real but equal, and the question asks for distinct ones. This is the sub-part that separates a careful answer from a careless one.

(iv) $3x^2 – 4x + 1 = 0$

$D = 16 – 12 = 4 > 0$ — Yes.

(v) $(x+4)^2 – 8x = 0$

Simplify first: $x^2 + 8x + 16 – 8x = x^2 + 16$. So $D = 0 – 64 = -64 < 0$ — No.

(vi) $\left(x – \sqrt2\right)^2 – 2(x+1) = 0$

Expanding, $x^2 – 2\sqrt2 x + 2 – 2x – 2 = x^2 – \left(2\sqrt2 + 2\right)x$. The constant term cancels exactly, so

$$D = \left(2 + 2\sqrt2\right)^2 – 0 = \left(2 + 2\sqrt2\right)^2 > 0$$

— Yes. (Numerically $12 + 8\sqrt2 \approx 23.3$.) Since $c = 0$, the roots are $0$ and $2 + 2\sqrt2$, plainly distinct.

(vii) $\sqrt2 x^2 – \dfrac{3}{\sqrt2}x + \dfrac{1}{\sqrt2} = 0$

$$D = \frac{9}{2} – 4\left(\sqrt2\right)\left(\frac{1}{\sqrt2}\right) = \frac92 – 4 = \frac12 > 0$$

— Yes.

A note on the printed answer. The key gives $D = 1$ here, not $\frac12$. Both are right, for different forms of the same equation: multiplying through by $\sqrt2$ gives $2x^2 – 3x + 1 = 0$, whose discriminant is $9 – 8 = 1$. Scaling an equation by $k$ multiplies its discriminant by $k^2$, and $\frac12 \times \left(\sqrt2\right)^2 = 1$. The roots are unchanged and so is the sign of $D$, which is all the question turns on — but it is worth knowing that the number itself depends on which form you write the equation in.

(viii) $x(1-x) – 2 = 0$

Expand: $x – x^2 – 2 = 0$, that is $x^2 – x + 2 = 0$. $D = 1 – 8 = -7 < 0$ — No.

(ix) $(x-1)(x+2) + 2 = 0$

Expand: $x^2 + x – 2 + 2 = x^2 + x$. $D = 1 – 0 = 1 > 0$ — Yes (the roots are $0$ and $-1$).

(x) $(x+1)(x-2) + x = 0$

Expand: $x^2 – x – 2 + x = x^2 – 2$. $D = 0 + 8 = 8 > 0$ — Yes (the roots are $\pm\sqrt2$).

Exercise 4.2, Question 2

Write whether the following statements are true or false. Justify your answers.

(i) Every quadratic equation has exactly one root.

False. $x^2 = 1$ has two roots, $1$ and $-1$. A quadratic has exactly one root only in the special case of a repeated root, such as $x^2 = 0$.

(ii) Every quadratic equation has at least one real root.

False. $x^2 + 1 = 0$ has none, since $x^2 \ge 0$ for every real $x$ and so $x^2 + 1 \ge 1$.

(iii) Every quadratic equation has at least two roots.

False. The same example serves: $x^2 + 1 = 0$ has no real roots, so it certainly does not have two. (In Class 10, “root” means real root throughout.)

(iv) Every quadratic equation has at most two roots.

True. A quadratic polynomial has at most two zeroes, because a polynomial of degree $n$ has at most $n$ zeroes — three distinct roots would make $(x-p)(x-q)(x-r)$ a factor, which is of degree three.

(v) If the coefficient of $x^2$ and the constant term of a quadratic equation have opposite signs, then the quadratic equation has real roots.

True. Opposite signs means $ac < 0$, so $-4ac > 0$ and

$$D = b^2 – 4ac \ge 0 + (\text{positive}) > 0$$

since $b^2 \ge 0$ always. The roots are real, and in fact distinct.

(vi) If the coefficient of $x^2$ and the constant term have the same sign and if the coefficient of $x$ is zero, then the quadratic equation has no real roots.

True. With $b = 0$ the discriminant is $D = -4ac$, and $a$, $c$ of the same sign gives $ac > 0$, so $D = -4ac < 0$.

The two parts together are a neat pair: (v) says $ac < 0$ guarantees real roots whatever $b$ is, and (vi) says $ac > 0$ with $b = 0$ guarantees none.

Exercise 4.2, Question 3

A quadratic equation with integral coefficients has integral roots. Justify your answer.

Solution. False — integral coefficients guarantee nothing about the roots.

A single counterexample settles it. Take

$$x^2 – 3x + 1 = 0$$

whose coefficients $1$, $-3$, $1$ are all integers. Its discriminant is $9 – 4 = 5$, so the roots are

$$x = \frac{3 \pm \sqrt5}{2}$$

which are irrational, let alone non-integral.

Exercise 4.2, Question 4

Does there exist a quadratic equation whose coefficients are rational but both of its roots are irrational? Justify your answer.

Solution. Yes. What is needed is a discriminant that is positive but not a perfect square, because then $\sqrt{D}$ is irrational and so are both roots $\dfrac{-b \pm \sqrt D}{2a}$.

$$x^2 – 6x + 7 = 0 \implies D = 36 – 28 = 8$$

and $\sqrt8 = 2\sqrt2$ is irrational, giving roots

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

Both irrational, as required. Note they come as a conjugate pair — with rational coefficients, irrational roots always do.

Exercise 4.2, Question 5

Does there exist a quadratic equation whose coefficients are all distinct irrationals but both the roots are rationals? Why?

Solution. Yes. The trick is to take an equation with rational roots and multiply throughout by an irrational number — which changes every coefficient but no root.

Start from $x^2 – 7x + 12 = 0$, whose roots are $3$ and $4$. Multiplying by $\sqrt3$:

$$\sqrt3\,x^2 – 7\sqrt3\,x + 12\sqrt3 = 0$$

The coefficients $\sqrt3$, $-7\sqrt3$ and $12\sqrt3$ are all irrational and all distinct, and the roots are still $3$ and $4$, because multiplying an equation by a non-zero constant never changes its solutions.

Exercise 4.2, Question 6

Is $0.2$ a root of the equation $x^2 – 0.4 = 0$? Justify.

Solution. No. Substituting:

$$(0.2)^2 – 0.4 = 0.04 – 0.4 = -0.36 \ne 0$$

The tempting mistake is to read $0.2$ as “the square root of $0.4$” because $2$ and $4$ look related. They are not: $\sqrt{0.4} \approx 0.632$, and the actual roots of the equation are $\pm\sqrt{0.4}$.

Exercise 4.2, Question 7

If $b = 0$, $c < 0$, is it true that the roots of $x^2 + bx + c = 0$ are numerically equal and opposite in sign? Justify.

Solution. Yes. With $b = 0$ the equation is simply

$$x^2 + c = 0 \implies x^2 = -c$$

and since $c < 0$, $-c$ is positive, so a real square root exists:

$$x = \pm\sqrt{-c}$$

The two roots are $+\sqrt{-c}$ and $-\sqrt{-c}$ — the same size, opposite signs.

Both conditions are needed. Without $b = 0$ the roots would not be symmetric about zero; without $c < 0$ there would be no real roots at all. For instance $c = -4$ gives $x^2 = 4$ and roots $\pm 2$.

Common mistakes

• Classifying before simplifying. Exercise 4.1 Questions 1 and 2 are entirely about this: three “quadratics” in Question 1 collapse to linear equations, and the quartic-looking option (D) in Question 2 is genuinely quadratic.
• Missing that $k+1 = 0$ in Exercise 4.1 Question 1(C). Substituting $k = -1$ removes the $x^2$ term altogether. A coefficient containing a letter always needs checking against the given value.
• Discarding $k = 0$ in Exercise 4.1 Question 6. At $k = 0$ the equation is $2x^2 = 0$, still quadratic (the leading coefficient is $2$), with the equal roots $0$ and $0$. Both roots of $k(k-8) = 0$ count.
• Confusing $c$ with $c^2$ in Exercise 4.1 Question 7. The constant added and subtracted is the square of the half-coefficient, $\frac{1}{64}$; option (A) offers $\frac18$, which is the half-coefficient itself.
• Reading “$D = 0$” as “two distinct real roots” in Exercise 4.2 Question 1(iii). The roots are real and equal. “Distinct” requires $D > 0$ strictly.
• Estimating $12\sqrt2$ and $8\sqrt2$ carelessly in Exercise 4.1 Questions 9(C) and 10(A). Both discriminants are close to zero — $-2.31$ and $-0.97$ — so a rough $\sqrt2 \approx 1.4$ is not always safe; use $1.414$.
• Expecting the discriminant to be a fixed number for an equation. Exercise 4.2 Question 1(vii) shows it is not: scaling the equation by $k$ scales $D$ by $k^2$. The sign, and therefore the conclusion, never changes.
• Reading $0.2$ as $\sqrt{0.4}$ in Exercise 4.2 Question 6. $(0.2)^2 = 0.04$, one decimal place further along than $0.4$.

Practise next

• Exemplar Exercises 4.3 and 4.4 — the same discriminant test followed through to the actual roots, plus the word problems.
• NCERT textbook Exercise 4.3 — the nature of the roots as the book sets it, which is Exercise 4.2 Question 1 above in its board-paper form.
• NCERT textbook Exercise 4.2 for solution by factorisation, and Exercise 4.1 for recognising and forming quadratic equations, which is Questions 1 and 2 above.
• Exemplar Exercises 2.1 and 2.2 — the sum-and-product relations used in Question 5, from the polynomial side.