NCERT Class 10 Mathematics — Quadratic Equations, Exercise 4.3. All 5 questions solved.

Exercise 4.3 is about predicting what the roots of a quadratic will be like before finding them. The tool is the quantity under the square root in the quadratic formula. For $ax^2 + bx + c = 0$ with $a \ne 0$,

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

and the discriminant $D = b^2 – 4ac$ decides the nature of the roots:

• $D > 0$: two distinct real roots;
• $D = 0$: two equal real roots, both $x = -\dfrac{b}{2a}$;
• $D < 0$: no real roots, since no real number has a negative square.

Key insight. The discriminant answers “are there real roots?” without solving anything. That turns every “is it possible?” question in this exercise into the same three steps: model the situation as a quadratic, compute $D$, and read its sign. And “equal roots” becomes the equation $D = 0$, which is how question 2 finds $k$.

Question 1

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

(i) $2x^2 – 3x + 5 = 0$    (ii) $3x^2 – 4\sqrt{3}\,x + 4 = 0$    (iii) $2x^2 – 6x + 3 = 0$

Solution. Compute $D$ first. Only if it is non-negative is it worth reaching for the formula.

(i) Here $a = 2$, $b = -3$, $c = 5$:

$$D = (-3)^2 – 4(2)(5) = 9 – 40 = -31 < 0$$

So the equation has no real roots.

(ii) Here $a = 3$, $b = -4\sqrt{3}$, $c = 4$. Squaring $b$ squares both the $4$ and the $\sqrt{3}$:

$$D = (-4\sqrt{3})^2 – 4(3)(4) = 48 – 48 = 0$$

So the roots are real and equal, each

$$x = -\frac{b}{2a} = \frac{4\sqrt{3}}{6} = \frac{2\sqrt{3}}{3} = \frac{2}{\sqrt{3}}$$

In fact the equation is the perfect square $(\sqrt{3}\,x – 2)^2 = 0$, which is why the two roots coincide.

(iii) Here $a = 2$, $b = -6$, $c = 3$:

$$D = (-6)^2 – 4(2)(3) = 36 – 24 = 12 > 0$$

So there are two distinct real roots:

$$x = \frac{6 \pm \sqrt{12}}{4} = \frac{6 \pm 2\sqrt{3}}{4} = \frac{3 \pm \sqrt{3}}{2}$$

Question 2

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

(i) $2x^2 + kx + 3 = 0$    (ii) $kx(x – 2) + 6 = 0$

Solution. Two equal roots means exactly $D = 0$, so set the discriminant to zero and solve for $k$.

(i) Here $a = 2$, $b = k$, $c = 3$:

$$k^2 – 4(2)(3) = 0 \quad\Longrightarrow\quad k^2 = 24 \quad\Longrightarrow\quad k = \pm\sqrt{24} = \pm 2\sqrt{6}$$

Both signs work: with $k = 2\sqrt{6}$ the equation is $(\sqrt{2}\,x + \sqrt{3})^2 = 0$, and with $k = -2\sqrt{6}$ it is $(\sqrt{2}\,x – \sqrt{3})^2 = 0$.

(ii) The equation is not yet in standard form, and $a$, $b$, $c$ cannot be read off until it is:

$$kx^2 – 2kx + 6 = 0, \qquad a = k,\ b = -2k,\ c = 6$$

Then

$$D = (-2k)^2 – 4(k)(6) = 4k^2 – 24k = 4k(k – 6) = 0$$

so $k = 0$ or $k = 6$. But $k = 0$ wipes out the $x^2$ term and leaves $6 = 0$, which is not a quadratic equation at all (and has no roots). So $k = 6$. Check: $6x^2 – 12x + 6 = 6(x – 1)^2$, whose roots are $1, 1$.

Question 3

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.

Solution. Let the breadth be $x$ m; then the length is $2x$ m and

$$x \cdot 2x = 800 \quad\Longrightarrow\quad 2x^2 – 800 = 0$$

With $a = 2$, $b = 0$, $c = -800$, the discriminant is

$$D = 0^2 – 4(2)(-800) = 6400 > 0$$

so real roots exist and the design is possible. The roots are $x = \pm\sqrt{400} = \pm 20$. A breadth cannot be negative, so $x = 20$: the breadth is $20$ m and the length $40$ m. Check: $40 \times 20 = 800$.

Question 4

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 $48$.

Solution. Let one friend’s present age be $x$ years; the other is $20 – x$. Four years ago both were $4$ years younger: $x – 4$ and $16 – x$. So

$$(x – 4)(16 – x) = 48 \quad\Longrightarrow\quad -x^2 + 20x – 64 = 48 \quad\Longrightarrow\quad x^2 – 20x + 112 = 0$$

Now test for real roots:

$$D = (-20)^2 – 4(1)(112) = 400 – 448 = -48 < 0$$

There is no real value of $x$, so no pair of ages fits both conditions. (This fits a quick check. Four years ago the two ages added to $12$, and the largest product two numbers with sum $12$ can have is $6 \times 6 = 36$, which is less than $48$.)

Question 5

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.

Solution. The perimeter is $2(l + b) = 80$, so $l + b = 40$. Let the length be $x$ m and the breadth $(40 – x)$ m. The area condition gives

$$x(40 – x) = 400 \quad\Longrightarrow\quad x^2 – 40x + 400 = 0$$

$$D = (-40)^2 – 4(1)(400) = 1600 – 1600 = 0$$

So there is exactly one (repeated) root, $x = \dfrac{40}{2} = 20$. The length is $20$ m and the breadth $40 – 20 = 20$ m. The design is possible, and the park turns out to be a square. A square is a rectangle, so this is a valid answer.

Common mistakes

• Question 1(ii), squaring only part of $-4\sqrt{3}$. $(-4\sqrt{3})^2$ is $16 \times 3 = 48$, and the minus sign disappears. Writing $4 \times 3 = 12$ (forgetting to square the $4$) gives $D = -36$ and a wrong verdict of “no real roots”.
• Question 1(i), writing “no roots”. The correct statement is no real roots. The equation does have roots in a larger number system, as Class 11 shows, so the word “real” is doing real work here.
• Question 2(i), giving only $k = 2\sqrt{6}$. $k^2 = 24$ has two solutions, and both give equal roots. Dropping the negative one loses half the answer.
• Question 2(ii), reading $a$, $b$, $c$ before expanding. Until $kx(x – 2) + 6$ is written as $kx^2 – 2kx + 6$, the coefficient of $x$ is not visible. And once $D = 4k(k – 6)$ appears, $k = 0$ must be rejected because it destroys the quadratic.
• Question 4, subtracting $4$ from one age only. “Four years ago” applies to both friends. Using $(x – 4)(20 – x) = 48$ instead gives $x^2 – 24x + 128 = 0$ with roots $8$ and $16$. That answers “yes” to a question whose answer is “no”.
• Question 5, taking $l + b = 80$. The perimeter counts each side twice, so $l + b$ is half of it, $40$. Using $80$ gives $x^2 – 80x + 400 = 0$, which has real roots but describes a park with the wrong perimeter.

Practise next

• Exercise 4.2 — factorisation. Its question 1(iv) and (v) are the $D = 0$ case seen from the other side, with a perfect square giving a repeated root.
• Class 11 Complex Numbers and Quadratic Equations, Exercise 4.1 — complex numbers, the larger number system in which equations with $D < 0$, like question 1(i) here, do have roots.