NCERT Solutions Class 10 Maths Chapter 4: Quadratic Equations
Detailed solutions for Exercise 4.3 (Nature of Roots).
#### **Q1. Find the nature of the roots of the following quadratic equations. If the real roots exist, find them:**
**(i)** $2x^2 – 3x + 5 = 0$
**(ii)** $3x^2 – 4\sqrt{3}x + 4 = 0$
**(iii)** $2x^2 – 6x + 3 = 0$
#### **A1. Solution:**
We compare the given equations with $ax^2 + bx + c = 0$ to find the Discriminant $D = b^2 – 4ac$.
* If $D > 0$, there are two distinct real roots.
* If $D = 0$, there are two equal real roots.
* If $D < 0$, there are no real roots.
**(i) $2x^2 – 3x + 5 = 0$**
Here, $a = 2, b = -3, c = 5$.
$D = b^2 - 4ac = (-3)^2 - 4(2)(5)$
$D = 9 - 40 = -31$
Since $D < 0$, **no real roots exist.**
**(ii) $3x^2 – 4\sqrt{3}x + 4 = 0$**
Here, $a = 3, b = -4\sqrt{3}, c = 4$.
$D = b^2 - 4ac = (-4\sqrt{3})^2 - 4(3)(4)$
$D = 48 - 48 = 0$
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 are $\frac{2\sqrt{3}}{3}$ and $\frac{2\sqrt{3}}{3}$.**
**(iii) $2x^2 – 6x + 3 = 0$**
Here, $a = 2, b = -6, c = 3$.
$D = b^2 - 4ac = (-6)^2 - 4(2)(3)$
$D = 36 - 24 = 12$
Since $D > 0$, **two distinct real roots exist.**
Roots are given by quadratic formula: $x = \frac{-b \pm \sqrt{D}}{2a}$
$x = \frac{-(-6) \pm \sqrt{12}}{2(2)} = \frac{6 \pm 2\sqrt{3}}{4}$
$x = \frac{2(3 \pm \sqrt{3})}{4} = \frac{3 \pm \sqrt{3}}{2}$
**Roots are $\frac{3 + \sqrt{3}}{2}$ and $\frac{3 - \sqrt{3}}{2}$.**
#### **Q2. Find the values of $k$ for each of the following quadratic equations, so that they have two equal roots.**
**(i)** $2x^2 + kx + 3 = 0$
**(ii)** $k x (x – 2) + 6 = 0$
#### **A2. Solution:**
For equal roots, the Discriminant must be zero ($D = b^2 – 4ac = 0$).
**(i) $2x^2 + kx + 3 = 0$**
$a = 2, b = k, c = 3$.
$D = k^2 - 4(2)(3) = k^2 - 24$
For equal roots, $D = 0$:
$k^2 - 24 = 0$
$k^2 = 24$
$k = \pm\sqrt{24} = \pm 2\sqrt{6}$
**Answer: $k = 2\sqrt{6}$ or $k = -2\sqrt{6}$.**
**(ii) $k x (x – 2) + 6 = 0$**
Simplify: $kx^2 - 2kx + 6 = 0$.
$a = k, b = -2k, c = 6$.
$D = (-2k)^2 - 4(k)(6)$
$D = 4k^2 - 24k$
For equal roots, $D = 0$:
$4k^2 - 24k = 0$
$4k(k - 6) = 0$
$k = 0$ or $k = 6$.
Since $a = k$ cannot be 0 for a quadratic equation, we reject $k=0$.
**Answer: $k = 6$.**
#### **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.**
#### **A3. Solution:**
Let the breadth be $x$ meters.
Then, the length is $2x$ meters.
Area = Length $\times$ Breadth
$2x \cdot x = 800$
$2x^2 = 800$
$x^2 = 400$
$x^2 - 400 = 0$
Here $D = 0^2 - 4(1)(-400) = 1600 > 0$. **Yes, it is possible.**
$x = \sqrt{400} = 20$ (Length cannot be negative, so ignore -20).
Breadth = 20 m.
Length = $2 \times 20 = 40$ m.
**Answer: 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.**
#### **A4. Solution:**
Let the age of the first friend be $x$.
Then the age of the second friend is $20 - x$.
**Four years ago:**
Age of first friend = $x - 4$
Age of second friend = $(20 - x) - 4 = 16 - x$
**Product:**
$(x - 4)(16 - x) = 48$
$16x - x^2 - 64 + 4x = 48$
$-x^2 + 20x - 64 - 48 = 0$
$-x^2 + 20x - 112 = 0$
$x^2 - 20x + 112 = 0$
**Check Discriminant:**
$a = 1, b = -20, c = 112$
$D = (-20)^2 - 4(1)(112)$
$D = 400 - 448 = -48$
Since $D < 0$, no real roots exist.
**Answer: This 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.**
#### **A5. Solution:**
Let length be $l$ and breadth be $b$.
1. Perimeter $= 2(l + b) = 80 \Rightarrow l + b = 40 \Rightarrow l = 40 - b$.
2. Area $= l \times b = 400$.
Substitute $l$ in the area equation:
$(40 - b)b = 400$
$40b - b^2 = 400$
$b^2 - 40b + 400 = 0$
**Check Discriminant:**
$a = 1, b = -40, c = 400$
$D = (-40)^2 - 4(1)(400)$
$D = 1600 - 1600 = 0$
Since $D = 0$, equal real roots exist. **Yes, it is possible.**
$b = \frac{-(-40)}{2(1)} = \frac{40}{2} = 20$.
Breadth = 20 m.
Length = $40 - 20 = 20$ m.
(Since length equals breadth, it is a square).
**Answer: Length = 20 m, Breadth = 20 m.**