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Class 10 RD Sharma Solutions – Chapter 8 Quadratic Equations – Exercise 8.4

  • 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:

=> (ab)2 = a2−2×a×b+b2

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 


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