RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Exercise: 10.1

RS Aggarwal Class 10 Solutions Chapter 10 Exercise: 10.1 Quadratic Equations

All these RS Aggarwal class 10 solutions Chapter 10 Exercise: 10.1 Quadratic Equations are solved by Byju's top ranked professors as per CBSE guidelines.

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.

Related Links
NCERT Books NCERT Solutions RS Aggarwal
Lakhmir Singh RD Sharma Solutions NCERT Solutions Class 6 to 12
More RS Aggarwal Solutions
RS Aggarwal Solutions Class 10 Solutions Chapter 10 Quadratic Equations Exercise 10 5RS Aggarwal Solutions Class 10 Solutions Chapter 11 Arithmetic Progressions Exercise 11 1
RS Aggarwal Solutions Class 10 Solutions Chapter 11 Arithmetic Progressions Exercise 11 5RS Aggarwal Solutions Class 10 Solutions Chapter 13 Constructions Exercise 13 1
RS Aggarwal Solutions Class 10 Solutions Chapter 15 Probability Exercise 15 3RS Aggarwal Solutions Class 10 Solutions Chapter 16 Coordinate Geometry Exercise 16 4
RS Aggarwal Solutions Class 10 Solutions Chapter 18 Areas Of Circle Sector And Segment Exercise 18 1RS Aggarwal Solutions Class 10 Solutions Chapter 18 Areas Of Circle Sector And Segment Exercise 18 5