NCERT Class 10 Maths – Exercise 4.3 Solutions

NCERT Class 10 Maths

Chapter 4 – Quadratic Equations | Exercise 4.3

(Rationalized Syllabus 2025-26)

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

(i) $2x^2 – 3x + 5 = 0$

Here, $a = 2, b = -3, c = 5$.

Discriminant $D = b^2 – 4ac$

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

Since $D < 0$, no real roots exist.

✖️ No Real Roots

(ii) $3x^2 – 4\sqrt{3}x + 4 = 0$

Here, $a = 3, b = -4\sqrt{3}, c = 4$.

Discriminant $D = b^2 – 4ac$

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

Since $D = 0$, two equal real roots exist.

Roots are given by $x = \frac{-b}{2a}$:

$$x = \frac{-(-4\sqrt{3})}{2(3)} = \frac{4\sqrt{3}}{6} = \frac{2\sqrt{3}}{3}$$

✔️ Roots: $\frac{2\sqrt{3}}{3}, \frac{2\sqrt{3}}{3}$

(iii) $2x^2 – 6x + 3 = 0$

Here, $a = 2, b = -6, c = 3$.

Discriminant $D = b^2 – 4ac$

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

Since $D > 0$, distinct real roots exist.

Using Quadratic Formula:

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

✔️ Roots: $\frac{3+\sqrt{3}}{2}, \frac{3-\sqrt{3}}{2}$

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

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

For equal roots, Discriminant $D = 0$.

$$b^2 – 4ac = 0$$

$$k^2 – 4(2)(3) = 0$$

$$k^2 – 24 = 0 \Rightarrow k^2 = 24$$

$$k = \pm\sqrt{24} = \pm 2\sqrt{6}$$

✔️ k = $\pm 2\sqrt{6}$

(ii) $kx(x – 2) + 6 = 0$

Rewrite in standard form:

$$kx^2 – 2kx + 6 = 0$$

Here $a=k, b=-2k, c=6$. For equal roots, $D=0$:

$$(-2k)^2 – 4(k)(6) = 0$$

$$4k^2 – 24k = 0$$

$$4k(k – 6) = 0$$

So $k = 0$ or $k = 6$.

However, if $k=0$, the equation becomes $6=0$ which is false (and the $x^2$ term vanishes). So $k \neq 0$.

✔️ k = 6

Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800 m²? If so, find its length and breadth.

Let breadth $= x$ meters.

Then length $= 2x$ meters.

Area = Length × Breadth = $2x \cdot x = 2x^2$.

$$2x^2 = 800$$

$$x^2 = 400 \Rightarrow x = \pm 20$$

Since breadth cannot be negative, $x = 20$.

Breadth = 20 m, Length = 40 m.

✔️ Yes, possible. Length = 40 m, Breadth = 20 m

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.

Let the age of the first friend be $x$.

Then the age of the second friend is $(20 – x)$.

4 years ago:

Age of 1st friend $= x – 4$

Age of 2nd friend $= (20 – x) – 4 = 16 – x$

Product is 48:

$$(x – 4)(16 – x) = 48$$

$$16x – x^2 – 64 + 4x = 48$$

$$-x^2 + 20x – 112 = 0 \Rightarrow x^2 – 20x + 112 = 0$$

Check Discriminant ($D = b^2 – 4ac$):

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

Since $D < 0$, no real roots exist.

✖️ No, this situation is not possible.

Is it possible to design a rectangular park of perimeter 80 m and area 400 m²? If so, find its length and breadth.

Perimeter $= 2(l + b) = 80 \Rightarrow l + b = 40 \Rightarrow b = 40 – l$.

Area $= l \times b = 400$.

$$l(40 – l) = 400$$

$$40l – l^2 = 400 \Rightarrow l^2 – 40l + 400 = 0$$

Check Discriminant:

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

Since $D=0$, equal real roots exist.

$$l = \frac{-(-40)}{2} = 20$$

If length $= 20$ m, then breadth $= 40 – 20 = 20$ m.

✔️ Yes, possible. Length = 20 m, Breadth = 20 m (It is a square)

🎉 Exercise 4.3 Completed | Chapter 4 Finished!