Class 10 RD Sharma Solutions Chapter 8 Quadratic Equations Exercise 8.4 (original) (raw)
Last Updated : 30 Apr, 2021
Question 1: Find the roots of the following quadratic (if they exist) by the method of completing the square: x^2-4\sqrt{2x}+6=0 .
Solution:
Given: x^2-4\sqrt{2x}+6=0
We have to make the equation a perfect square.
=> x^2-4\sqrt{2x} = -6
=> x^2 - 2\times2\sqrt{2x} + (2\sqrt{2})^2 = -6 + (2\sqrt{2})^2
We know that:
=> (a_−_b)2 = a_2−2×_a_×_b+_b_2
Thus, the equation can be written as:
=> (x-2\sqrt{2})^2 = -6 + (2\sqrt{2})^2
=> (x-2\sqrt{2})^2 = -6 + (2)^2(\sqrt{2})^2
=> (x-2\sqrt{2})^2 = -6 + (4\times2)
=> (x-2\sqrt{2})^2 = -6 + 8
=> (x-2\sqrt{2})^2 = 2
The RHS is positive, which implies that the roots exist.
=> (x-2\sqrt{2}) = \pm \sqrt{2}
=> x = \sqrt{2}+2\sqrt{2} and x= -\sqrt{2}+2\sqrt{2}
=> x = 3\sqrt{2} and x = \sqrt{2}
Question 2: Find the roots of the following quadratic (if they exist) by the method of completing the square: 2x2-7x+3 = 0.
Solution:
Given: 2x2-7x+3 = 0
We have to make the equation a perfect square.
=> 2x2-7x+3 = 0
=> x^2 - \dfrac{7}{2}x+\dfrac{3}{2}=0
=> x^2 -2\times \dfrac{7}{4}\times x+(\dfrac{7}{4})^2 - (\dfrac{7}{4})^2+\dfrac{3}{2} = 0
We know that:
=> (a−b)2=a2−2×a×b+b2
Thus, the equation can be written as:
=> (x-\dfrac{7}{4})^2 -\dfrac{49}{16} + \dfrac{3}{2}=0
=> (x-\dfrac{7}{4})^2 - \dfrac{25}{16} = 0
=> (x-\dfrac{7}{4})^2 = \dfrac{25}{16}
The RHS is positive, which implies that the roots exist.
=> (x-\dfrac{7}{4})^2 = \pm \dfrac{5}{4}
=> x = \dfrac{7}{4}+\dfrac{5}{4} and x= \dfrac{7}{4}-\dfrac{5}{4}
=> x = \dfrac{12}{4} and x = \dfrac{2}{4}
=> x = 3 and x = \dfrac{1}{2}
Question 3: Find the roots of the following quadratic (if they exist) by the method of the completing the square: 3x2+11x+10 = 0.
Solution:
Given: 3x2+11x+10 = 0
We have to make the equation a perfect square.
=> 3x2+11x+10 = 0
=> x^2 + \dfrac{11}{3} x +\dfrac{10}{3}=0
=> x^2 + 2(\dfrac{11}{6})x+(\dfrac{11}{6})^2-(\dfrac{11}{6})^2+\dfrac{10}{3}=0
We know that:
=> (a−b)2=a2−2×a×b+b2
Thus the equation can be written as:
=> (x+\dfrac{11}{6})^2+\dfrac{10}{3}-\dfrac{121}{36}=0
=> (x+\dfrac{11}{6})^2 = \dfrac{1}{36}
The RHS is positive, which implies that the roots exist.
=> (x+\dfrac{11}{6})= \pm \dfrac{1}{6}
=> x = -\dfrac{11}{6}+\dfrac{1}{6} and x= -\dfrac{11}{6}-\dfrac{1}{6}
=> x = -\dfrac{10}{6} and x = -\dfrac{12}{6}
=> x = -\dfrac{5}{3} and x = -2
Question 4: Find the roots of the following quadratic (if they exist) by the method of completing the square: 2x2+x-4 =0.
Solution:
Given: 2x2+x-4 =0
We have to make the equation a perfect square.
=> 2x2+x-4 =0
=> x^2 - \dfrac{x}{2}-2=0
=> x^2 + 2\times \dfrac{1}{4}\times x+(\dfrac{1}{4})^2-(\dfrac{1}{4})^2-2=0
We know that:
=> (a−b)2=a2−2×a×b+b2
Thus the equation can be written as:
=> (x+\dfrac{1}{4})^2 = \dfrac{33}{16}
The RHS is positive, which implies that the roots exist.
=> (x+\dfrac{1}{4})= \pm \dfrac{\sqrt{33}}{4}
=> x = \dfrac{\sqrt{33}-1}{4} and x= \dfrac{-\sqrt{33}-1}{4}
Question 5: Find the roots of the following quadratic (if they exist) by the method of completing the square: 2x2+x+4 =0.
Solution:
Given: 2x2+x+4 =0
We have to make the equation a perfect square.
=> 2x2+x+4 =0
=> x^2 + \dfrac{x}{2}+2=0
=> x^2 + 2\times \dfrac{1}{4}\times x+(\dfrac{1}{4})^2-(\dfrac{1}{4})^2+2=0
We know that:
=> (a−b)2=a2−2×a×b+b2
Thus the equation can be written as:
=> (x+\dfrac{1}{4})^2 = -\dfrac{31}{16}
=> The RHS is negative, which implies that the roots are not real.
Question 6: Find the roots of the following quadratic (if they exist) by the method of completing the square: 4x2+4√3+3=0.
Solution:
Given: 4x2+4√3+3=0
We have to make the equation a perfect square.
=> 4x2+4√3+3=0
=> x^2 + \sqrt{3}x+\dfrac{3}{4} =0
=> x^2 + 2\times x \times \dfrac{\sqrt{3}}{2} + (\dfrac{\sqrt{3}}{2})^2 - (\dfrac{\sqrt{3}}{2})^2+ \dfrac{3}{4}=0
We know that,
=> (a−b)2=a2−2×a×b+b2
Thus the equation can be written as:
=> (x+\dfrac{\sqrt{3}}{2})^2 = -\dfrac{3}{4}+(\dfrac{\sqrt{3}}{2})^2
=> (x+\dfrac{\sqrt{3}}{2})^2 = -\dfrac{3}{4}+\dfrac{3}{4}
=> (x+\dfrac{\sqrt{3}}{2})^2 = 0
The RHS is zero, which implies that the roots exist and are equal.
=> x = -\dfrac{\sqrt{3}}{2}
Question 7: Find the roots of the following quadratic (if they exist) by the method of the completing the square: \sqrt{2}x^2 -3x-2\sqrt{2}=0 .
Solution:
Given: \sqrt{2}x^2 -3x-2\sqrt{2}=0
We have to make the equation a perfect square.
=> \sqrt{2}x^2 -3x-2\sqrt{2}=0
=> x^2 -\dfrac{3}{\sqrt{2}}x-2 =0
=> x^2 - 2\times x \times \dfrac{3}{2\sqrt{2}} + (\dfrac{3}{2\sqrt{2}})^2 - (\dfrac{3}{2\sqrt{2}})^2-2=0
We know that,
=> (a−b)2=a2−2×a×b+b2
Thus the equation can be written as:
=> (x-\dfrac{3}{2\sqrt{2}})^2 = 2 + (\dfrac{3}{2\sqrt{2}})^2
=> (x-\dfrac{3}{2\sqrt{2}})^2 = 2+ \dfrac{9}{8}
=> (x-\dfrac{3}{2\sqrt{2}})^2 = \dfrac{25}{8}
The RHS is positive, which implies that the roots exist.
=> (x-\dfrac{3}{2\sqrt{2}}) = \dfrac{5}{2\sqrt{2}}
=> x = \dfrac{5+3}{2\sqrt{2}} and x = \dfrac{-5+3}{2\sqrt{2}}
=> x = 2\sqrt{2} and x = \dfrac{-1}{\sqrt{2}}
Question 8: Find the roots of the following quadratic (if they exist) by the method of completing the square: \sqrt{3}x^2+10x+7\sqrt{3}=0 .
Solution:
Given: \sqrt{3}x^2+10x+7\sqrt{3}=0
We have to make the equation a perfect square.
=> \sqrt{3}x^2+10x+7\sqrt{3}=0
=> x^2 +\dfrac{10}{\sqrt{3}}x+7 =0
=> x^2 +2\times x \times \dfrac{5}{\sqrt{3}} + (\dfrac{5}{\sqrt{3}})^2 - (\dfrac{5}{\sqrt{3}})^2+7=0
We know that,
=> (a−b)2=a2−2×a×b+b2
Thus the equation can be written as:
=> (x+\dfrac{5}{\sqrt{3}})^2 = -7 + (\dfrac{5}{\sqrt{3}})^2
=> (x+\dfrac{5}{\sqrt{3}})^2 = -7+ \dfrac{25}{3}
=> (x+\dfrac{5}{\sqrt{3}})^2 = \dfrac{4}{3}
The RHS is positive, which implies that the roots exist.
=> (x+\dfrac{5}{\sqrt{3}}) = \pm \dfrac{2}{\sqrt{3}}
=> x = \dfrac{-5+2}{\sqrt{3}} and x = \dfrac{-5-2}{\sqrt{3}}
=> x = -\sqrt{3} and x = \dfrac{-7}{\sqrt{3}}
Question 9: Find the roots of the following quadratic (if they exist) by the method of completing the square: x^2-(\sqrt{2}+1)x+\sqrt{2}=0 .
Solution:
Given: x^2-(\sqrt{2}+1)x+\sqrt{2}=0
We have to make the equation a perfect square.
=> x^2-(\sqrt{2}+1)x+\sqrt{2}=0
=> x^2-(\sqrt{2}+1)x+(\dfrac{\sqrt{2}+1}{2})^2 - (\dfrac{\sqrt{2}+1}{2})^2+\sqrt{2}=0
We know that,
=> (a−b)2=a2−2×a×b+b2
Thus the equation can be written as:
=> (x-(\dfrac{\sqrt{2}+1}{2}))^2 = \dfrac{2+1+2\sqrt{2}}{4}-\sqrt{2}
=> (x-(\dfrac{\sqrt{2}+1}{2}))^2 = \dfrac{2+1-2\sqrt{2}}{4}
=> (x-(\dfrac{\sqrt{2}+1}{2}))^2 = (\dfrac{\sqrt{2}-1}{2})^2
The RHS is positive, which implies that the roots exist.
=> (x-(\dfrac{\sqrt{2}+1}{2})) = \pm (\dfrac{\sqrt{2}-1}{2})
=> x = \dfrac{\sqrt{2}-1+\sqrt{2}+1}{2} and x = \dfrac{-\sqrt{2}+1+\sqrt{2}+1}{2}
=> x = √2 and x = 1
Question 10: Find the roots of the following quadratic equation (if they exist) by the method of completing the square: x2-4ax+4a2-b2=0.
Solution:
Given: x2-4ax+4a2-b2=0
We have to make the equation a perfect square.
=> x2-4ax+4a2-b2=0
=> x2−2×x×2a+(2a)2−b2=0
We know that,
=> (a−b)2=a2−2×a×b+b2
Thus the equation can be written as:
=> x2−2×2a×x+(2a)2=b2
=> (x-2a)2 = b2
The RHS is positive, which implies that the roots exist.
=> (x-2a) = ±b
=> x= 2a+b and x = 2a-b