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

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= 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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Last Updated : 30 Apr, 2021

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

Question 1: Find the roots of the following quadratic (if they exist) by the method of completing the square: $x^2-4\sqrt{2x}+6=0$ .

Question 2: Find the roots of the following quadratic (if they exist) by the method of completing the square: 2x²-7x+3 = 0.

Question 3: Find the roots of the following quadratic (if they exist) by the method of the completing the square: 3x²+11x+10 = 0.

Question 4: Find the roots of the following quadratic (if they exist) by the method of completing the square: 2x²+x-4 =0.

Question 5: Find the roots of the following quadratic (if they exist) by the method of completing the square: 2x²+x+4 =0.

Question 6: Find the roots of the following quadratic (if they exist) by the method of completing the square: 4x²+4√3+3=0.

Question 7: Find the roots of the following quadratic (if they exist) by the method of the completing the square: $\sqrt{2}x^2 -3x-2\sqrt{2}=0$ .

Question 8: Find the roots of the following quadratic (if they exist) by the method of completing the square: $\sqrt{3}x^2+10x+7\sqrt{3}=0$ .

Question 9: Find the roots of the following quadratic (if they exist) by the method of completing the square: $x^2-(\sqrt{2}+1)x+\sqrt{2}=0$ .

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.

Data Structures and Algorithms - Self Paced

CBSE Class 12 Computer Science

GATE CS & IT 2024

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

Question 1: Find the roots of the following quadratic (if they exist) by the method of completing the square: .

Question 2: Find the roots of the following quadratic (if they exist) by the method of completing the square: 2x2-7x+3 = 0.

Question 3: Find the roots of the following quadratic (if they exist) by the method of the completing the square: 3x2+11x+10 = 0.

Question 4: Find the roots of the following quadratic (if they exist) by the method of completing the square: 2x2+x-4 =0.

Question 5: Find the roots of the following quadratic (if they exist) by the method of completing the square: 2x2+x+4 =0.

Question 6: Find the roots of the following quadratic (if they exist) by the method of completing the square: 4x2+4√3​+3=0.

Question 7: Find the roots of the following quadratic (if they exist) by the method of the completing the square: .

Question 8: Find the roots of the following quadratic (if they exist) by the method of completing the square: .

Question 9: Find the roots of the following quadratic (if they exist) by the method of completing the square: .

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.

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Data Structures and Algorithms - Self Paced

CBSE Class 12 Computer Science

GATE CS & IT 2024

Question 1: Find the roots of the following quadratic (if they exist) by the method of completing the square: $x^2-4\sqrt{2x}+6=0$ .

Question 2: Find the roots of the following quadratic (if they exist) by the method of completing the square: 2x²-7x+3 = 0.

Question 3: Find the roots of the following quadratic (if they exist) by the method of the completing the square: 3x²+11x+10 = 0.

Question 4: Find the roots of the following quadratic (if they exist) by the method of completing the square: 2x²+x-4 =0.

Question 5: Find the roots of the following quadratic (if they exist) by the method of completing the square: 2x²+x+4 =0.

Question 6: Find the roots of the following quadratic (if they exist) by the method of completing the square: 4x²+4√3+3=0.

Question 7: Find the roots of the following quadratic (if they exist) by the method of the completing the square: $\sqrt{2}x^2 -3x-2\sqrt{2}=0$ .

Question 8: Find the roots of the following quadratic (if they exist) by the method of completing the square: $\sqrt{3}x^2+10x+7\sqrt{3}=0$ .

Question 9: Find the roots of the following quadratic (if they exist) by the method of completing the square: $x^2-(\sqrt{2}+1)x+\sqrt{2}=0$ .

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.