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.


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