# RS Aggarwal Solutions Class 10 Ex 10A

Q1: 3x2 — 2x — 1= 0

We write, —2x = —3x + x as 3x2 x (-1) = —3x2 = (-3x) x x

Therefore, 3x2 — 2x — 1= 0

$\Rightarrow$ 3x2 —3x + x – 1=0

$\Rightarrow$ 3x(x —1) -I-1(x — 1) = 0

$\Rightarrow$ (x —1)(3x +1) = 0

$\Rightarrow$ x — 1 = 0 or 3x +1= 0

$\Rightarrow$ x = 1 or x = –$\frac{1}{3}$

Hence, the roots of the given equation are 1 and — $\frac{1}{3}$.

Q2: 4x2 — 9x = 100

Given:

4x2 — 9x = 100

$\Rightarrow$ 4x2 — 9x — 100 = 0

$\Rightarrow$ 4×2 — (25x — 16x) — 100 = 0

$\Rightarrow$ 4x2 — 25x + 16x — 100 = 0

$\Rightarrow$ x(4x — 25) + 4(4x — 25) = 0

$\Rightarrow$ (4x — 25)(x + 4) = 0

$\Rightarrow$ 4x — 25 = 0 or x + 4 = 0

$\Rightarrow$ x= $\frac{25}{4}$ or x = -4

Hence, the roots of the equation are $\frac{25}{4}$ and — 4.

Q3: 15x2 — 28 = x

Given:

15x2 — 28 = x

$\Rightarrow$ 15x2 — x — 28 = 0

$\Rightarrow$ 15x2 — (21x — 20x) — 28 = 0

$\Rightarrow$ 15x2 — 21x + 20x — 28 = 0

$\Rightarrow$ 3x(5x — 7) + 4(5x — 7) = 0

$\Rightarrow$ (3x + 4)(5x — 7) = 0

$\Rightarrow$ 3x + 4 = 0 or 5x — 7 = 0

$\Rightarrow$ x= $\frac{-4}{3}$ or x = $\frac{7}{5}$

Hence, the roots of the equation are $\frac{-4}{3}$ or x = $\frac{7}{5}$ .

Q4: 4 — 11x = 3x2

Given:

4 — 11x = 3x2

$\Rightarrow$ 3x2 + 11x — 4 = 0

$\Rightarrow$ 3x2 + 12x — x — 4 = 0

$\Rightarrow$ 3x(x + 4) — 1(x + 4) = 0

$\Rightarrow$ (x + 4)(3x — 1) = 0

$\Rightarrow$ x + 4 = 0 or 3x — 1 = 0

$\Rightarrow$ x = —4 or x = $\frac{1}{3}$

Hence, the roots of the equation are — 4 and $\frac{1}{3}$ .

Q5: 48x2 — 13x — 1 = 0

Given:

48x2 — 13x — 1 = 0

$\Rightarrow$ 48x2 — (16x — 3x) — 1 = 0

$\Rightarrow$ 48x2 — 16x + 3x — 1 = 0

$\Rightarrow$ 16x(3x — 1) + 1(3x — 1) = 0

$\Rightarrow$ (16x + 1)(3x — 1) = 0

$\Rightarrow$ 16x + 1 = 0 or 3x — 1 =0

$\Rightarrow$ x = $\frac{-1}{16}$ or x = $\frac{1}{3}$

Hence, the roots of the equation are $\frac{-1}{16}$ and $\frac{1}{3}$ .

Q6: x2 + 2$\sqrt{2x}$– 6 = 0

We write, 2$\sqrt{2x}$= 3$\sqrt{2x}$$\sqrt{2x}$ as x2 x (-6) = —6x2 = 3$\sqrt{2x}$ x (-$\sqrt{2x}$)

Therefore, X2 + 2$\sqrt{2x}$– 6 = 0

$\Rightarrow$ x2+3$\sqrt{2x}$$\sqrt{2x}$-6=0

$\Rightarrow$ x(x+3$\sqrt{2}$) —$\sqrt{2}$ (x+3$\sqrt{2}$) =0

$\Rightarrow$ (x+3$\sqrt{2}$)(x—$\sqrt{2}$)=0

$\Rightarrow$ x+3$\sqrt{2}$=0 or x-$\sqrt{2}$=0

$\Rightarrow$ x= —3$\sqrt{2}$ or x=$\sqrt{2}$

Hence, the roots of the given equation are —3$\sqrt{2}$ and $\sqrt{2}$

Q7: $\sqrt{3}$x2 x 7$\sqrt{3}$

We write, 10x = 3x +7x as $\sqrt{3}$x2 x 7$\sqrt{3}$ = 21x2 = 3x x 7x

Therefore, $\sqrt{3}$x2+10x+7$\sqrt{3}$=0

$\Rightarrow$ $\sqrt{3}$x2+3x+7x+7$\sqrt{3}$=0

$\Rightarrow$ $\sqrt{3}$x(x +$\sqrt{3}$) +7(x +$\sqrt{3}$) =0

$\Rightarrow$ (x+ $\sqrt{3}$ )( $\sqrt{3}$x +7) = 0

$\Rightarrow$ x+$\sqrt{3}$=0 or $\sqrt{3}$x+7=0

$\Rightarrow$ x = –$\sqrt{3}$ or x = – $\frac{7}{\sqrt{3}}$ =- $\frac{7\sqrt{3}}{3}$

Hence, the roots of the given equation are —$\sqrt{3}$ and –$\frac{7\sqrt{3}}{3}$

Q8: $\sqrt{3}$x2 + 11x + 6$\sqrt{3}$

Given :

$\sqrt{3}$x2 + 11x + 6$\sqrt{3}$ = 0

$\Rightarrow$ $\sqrt{3}$x2 + 9x + 2x + 6$\sqrt{3}$ =0

$\Rightarrow$ $\sqrt{3x}$x(x + 3$\sqrt{3}$) + 2(x + 3$\sqrt{3}$) =0

$\Rightarrow$ (x + 3$\sqrt{3}$) ($\sqrt{3x}$ + 2) =0

$\Rightarrow$ x + 3$\sqrt{3}$ = 0 or $\sqrt{3}$x + 2 = 0

$\Rightarrow$ x = -3$\sqrt{3}$ or x = $\frac{-2}{\sqrt{3}}$ = $\frac{2\sqrt{3}}{\sqrt{3}\sqrt{3}}$ = $\frac{-2\sqrt{3}}{3}$

Hence, the roots of the equation are -3$\sqrt{3}$ and $\frac{-2\sqrt{3}}{3}$.

Q9: 3$\sqrt{7}$x2 + 4x — $\sqrt{7}$ = 0

Given:

3$\sqrt{7}$x2 + 4x — $\sqrt{7}$ = 0

$\Rightarrow$ 3$\sqrt{7}$ x2 + 7x — 3x — $\sqrt{7}$ = 0

$\Rightarrow$ $\sqrt{7}$ (3x + $\sqrt{7}$) – 1(3x + $\sqrt{7}$) = 0

$\Rightarrow$ (3x + $\sqrt{7}$) ($\sqrt{7}$ x – 1) = 0

$\Rightarrow$ 3x+$\sqrt{7}$ =0 or $\sqrt{7}$ x – 1 = 0

$\Rightarrow$ x = $\frac{-\sqrt{7}}{3}$ or x = $\frac{1}{\sqrt{7}}$ = $\frac{1 \times \sqrt{7}}{\sqrt{7} \times\sqrt{7}}$= $\frac{\sqrt{7}}{7}$

Hence, the roots of the equation are –$\frac{\sqrt{7}}{3}$

and $\frac{\sqrt{7}}{7}$.

Q10. $3x{2} – 2 \sqrt{6} x + 2 = 0$

We write, $-2 \sqrt{6} x = – \sqrt{6} x – sqrt{6} x as 3x{2} \times 2 = (- \sqrt{6} x ) \times (- \sqrt{6} x )$

Hence,

$3x{2} – 2 \sqrt{6} x + 2 = 0$

$\Rightarrow 3x{2} – \sqrt{6} x – \sqrt{6} x + 2 = 0$

$\Rightarrow \sqrt{3} x (\sqrt{3} x – \sqrt{2} ) – \sqrt{2} ( \sqrt{3} x – \sqrt{2} ) = 0$

$\Rightarrow ( \sqrt{3} x – \sqrt{2} ) ( \sqrt{3} x – \sqrt{2} ) = 0$

$\Rightarrow ( \sqrt{3} x – \sqrt{2} ){2} = 0$

$\Rightarrow \sqrt{3} x – \sqrt{2} = 0$

$\Rightarrow x = \frac{\sqrt{2}}{\sqrt{3}} = \frac{\sqrt{6}}{3}$

Hence, $\frac{\sqrt{6}}{3}$ is the represented root of the equation.

Q11: $\sqrt{3} x{2} – 2 \sqrt{2} x – 2 \sqrt{3} = 0$

We write, $– 2 \sqrt{2} x = -3 \sqrt{2} x + \sqrt{2} x as \sqrt{3}x{2} \times (-2 \sqrt{3} ) = -6x{2} = ( -3 \sqrt{2} x) \times (\sqrt{2} x)$

Hence,

$\sqrt{3} x{2} – 2 \sqrt{2} x – 2 \sqrt{3} = 0$

$\Rightarrow \sqrt{3} x{2} – 3 \sqrt{2} x + \sqrt{2} x – 2 \sqrt{3} = 0$

$\Rightarrow \sqrt{3} x (x – \sqrt{6} ) + \sqrt{2} (x – \sqrt{6} ) = 0$

$\Rightarrow (x – \sqrt{6}) (\sqrt{3} x + \sqrt{2}) = 0$

$\Rightarrow x – \sqrt{6} = 0 or \sqrt{3} x + \sqrt{2} = 0$

$\Rightarrow x = \sqrt{6} or x = – \frac{\sqrt{2}}{\sqrt{3}} = – \frac{\sqrt{6}}{3}$

Hence, the roots of the given equation are $\sqrt{6} and – \frac{\sqrt{6}}{3}$

Q12: $x{2} – 3 \sqrt{5} x + 10 = 0$

We write, $-3 \sqrt{5} x = -2 \sqrt{5} x – \sqrt{5} x as x{2} \times 10 x{2} = (-2 \sqrt{5} x ) \times (- \sqrt{5} x)$

Hence,

$x{2} – 3 \sqrt{5} x + 10 = 0$

$\Rightarrow x{2} – 2 \sqrt{5} x – \sqrt{5} x + 10 = 0$

$\Rightarrow x(x – 2\sqrt{5}) – \sqrt{5}(x – 2 \sqrt{5}) = 0$

$\Rightarrow (x – 2\sqrt{5})(x – \sqrt{5}) = 0$

$\Rightarrow x – \sqrt{5} = 0 or x – 2 \sqrt{5} = 0$

$\Rightarrow x = \sqrt{5} or x = 2 \sqrt{5}$

Hence, the roots of the given equations are $\sqrt{5} and 2 \sqrt{5}$

Q13: $x{2} – ( \sqrt{3} + 1) x + \sqrt{3} = 0$

$x{2} – ( \sqrt{3} + 1) x + \sqrt{3} = 0$

$\Rightarrow x{2} – \sqrt{3} x – x + \sqrt{3} = 0$

$\Rightarrow x(x – \sqrt{3}) – 1(x – \sqrt{3} ) = 0$

$\Rightarrow (x – \sqrt{3} )(x – 1) = 0$

$\Rightarrow x – \sqrt{3} = 0 or x = 1$

Hence, 1 and $\sqrt{3}$ are the roots of the given equation.

Q14: $x{2} + 3 \sqrt{3} x – 30 = 0$

We write,

$3\sqrt{3} x = 5 \sqrt{3} x – 2\sqrt{3} x as x{2} \times (-30) = -30x{2} = 5 \sqrt{3}x \times (-2 \sqrt{3}x)$

$x{2} + 3 \sqrt{3} x – 30 = 0$

$\Rightarrow x{2} + 5\sqrt{3}x – 2\sqrt{3} x – 30 = 0$

$\Rightarrow x(x + 5 \sqrt{3} ) – 2 \sqrt{3} x – 30 = 0$

$\Rightarrow (x + 5 \sqrt{3})(x – 2\sqrt{3}) = 0$

$\Rightarrow x + 5\sqrt{3} = 0 or x – 2\sqrt{3} = 0$

$\Rightarrow x = -5 \sqrt{3} or x = 2 \sqrt{3}$

Hence, the roots of the given equation are, $-5 \sqrt{3} and 2 \sqrt{3}$

Q15: $\sqrt{2}x{2} + 7x + 5\sqrt{2} = 0$

We write, 7x = 5x + 2x as $\sqrt{2} x{2} \times 5 \sqrt{2} = 10 x{2} = 5x \times 2x$

Hence,

$\sqrt{2}x{2} + 7x + 5\sqrt{2} = 0$

$\Rightarrow \sqrt{2} x{2} + 5x + 2x + 5\sqrt{2} = 0$

$\Rightarrow x(\sqrt{2}x + 5) + \sqrt{2} (\sqrt{2}x + 5 ) = 0$

$\Rightarrow(\sqrt{2} x + 5 )(x + \sqrt{2} ) = 0$

$\Rightarrow x + \sqrt{2} = 0 ot \sqrt{2} x + 5 = 0$

$\Rightarrow x = – \sqrt{2} or x = – \frac{5}{\sqrt{2}} = – \frac{5 \sqrt{2}}{2}$

Hence, the roots of the given equation are $- \sqrt{2} \; and \; \frac{-5 \sqrt{2}}{2}$

Q16: 5x2 + 13x + 8 = 0

We write 13x = 5x + 8x as 5x2 × 8 = 40x2 = 5x × 8 x

Hence,

5x2 + 13x + 8 = 0

$\Rightarrow$ 5x2 + 5x + 8x + 8 = 0

$\Rightarrow$ 5x(x + 1) + 8(x + 1) = 0

$\Rightarrow$ (x + 1)(5x + 8) = 0

$\Rightarrow$ x + 1 = 0 or 5x + 8 = 0

$\Rightarrow$ x = -1 and (-8/5) are the roots of the equation.

Q17: $x{2} – (1 + \sqrt{2}) x + \sqrt{2} = 0$

Given:

$x{2} – (1 + \sqrt{2}) x + \sqrt{2} = 0$

$\Rightarrow x{2} – x \sqrt{2}x + \sqrt{2} = 0$

$\Rightarrow x(x – 1) – \sqrt{2}(x – 1) = 0$

$\Rightarrow(x – \sqrt{2} )(x – 1) = 0$

$\Rightarrow x – \sqrt{2} = 0 or x – 1 = 0$

$\Rightarrow x = \sqrt{2} or x = 1$

Hence, the roots of the equation are $\sqrt{2} and 1$

Q18: 9x2 + 6x + 1 = 0

Given:

9x2 + 6x + 1 = 0

$\Rightarrow$ 9x2 + 3x + 3x + 1 = 0

$\Rightarrow$ 3x(3x + 1) + 1(3x + 1) = 0

$\Rightarrow$ (3x + 1)(3x + 1) = 0

$\Rightarrow$ 3x + 1 = 0 or 3x + 1 = 0

$\Rightarrow x = \frac{-1}{3} or x = \frac{-1}{3}$

Hence,

$\frac{-1}{3}$ is the root of the equation 9x2 + 6x + 1 = 0

Q19: 100x2 – 20x + 1 = 0

We write,

-20x = -10x – 10x as 100x2 × 1 = 100x2 = (-10x) × (-10x)

Hence,

100x2 – 20x + 1 = 0

$\Rightarrow$ 100x2 – 10x – 10x + 1 = 0

$\Rightarrow$ 10x(10x – 1) -1(10x – 1) = 0

$\Rightarrow$ (10x -1)(10x – 1) = 0

$\Rightarrow$ (10x – 1)2 = 0

$\Rightarrow$ 10x – 1 = 0

$\Rightarrow x = \frac{1}{10}$

Hence. $\frac{1}{10}$ is represented root of the given equation.

We write, -20x = -10x -10x as 100x2 x 1 = 100x2 = (-10x) x (-10x)

100x2 – 20x + 1= 0

100x2 – 10x – 10x + 1 = 0

10x(10x – 1) – 1(10x – 1) =0

(l0x – 1)(10x – 1) = 0 (10x – 1)2 = 0

10x – 1 = 0

10x=1

x=0.1 Hence, 0.1 is the repreated root of the given equation.

Q20: 2x2 –x +$\frac{1}{8}$ =0

We write –x= $\frac{-x}{2}$ $\frac{-x}{2}$ as $2x{2} \times\frac{1}{8}$

=$\frac{-x}{4}$ =$\frac{-x}{2} \times \frac{-1}{2}$

2x2 –x +$\frac{1}{8}$ =0

2x(x-$\frac{1}{4}$) $\frac{-1}{2}$ $\frac{x-1}{4}$

$x-\frac{1}{4}$ or x=$\frac{1}{4}$

Hence $\frac{1}{4}$ is the repeated root of the given equation

Q21: 10x- $\frac{1}{x}$ =3

Given:

10x- $\frac{1}{x}$ =3

10x2 – 1 = 3x [Multiplying both sides by x]

10x2 – 3x – 1 = 0

10x2 – (5x – 2x) – 1 = 0

10x2 – 5x + 2x – 1 = 0

5x(2x – 1) + 1(2x – 1) = 0

(2x – 1)(5x + 1) = 0

2x – 1 = 0 or 5x + 1 = 0

x = $\frac{1}{2}$ or x = $\frac{-1}{5}$

Hence, the roots of the equation are $\frac{1}{2}$ and $\frac{-1}{5}$

Q22: $\frac{2}{x{2}}$$\frac{5}{x}$ +2x =0

Given:

$\frac{2}{x{2}}$$\frac{5}{x}$ +2x=0

2 – 5x + 2x2 = 0 [Multiplying both side by x2]

2x2 – 5x + 2 = 0

2x2 – (4x + x) + 2 = 0

2x2 – 4x – x + 2 =0

2x(x – 2) – 1(x – 2) = 0

(2x – 1)(x – 2) = 0

2x – 1 = O or x – 2 = 0

X=$\frac{1}{2}$ or x=2

Hence, the roots of the equation are $\frac{1}{2}$ and 2.

#### Practise This Question

A(a,b), B(c,d) and C(e,f) are the vertices of a triangle.
i) AB + BC > AC
ii) Area of the triangle = ( 12)[a(c-e)+c(f-b)+e(b-d)]
Which of the following are true?