If $x = a^{2} + b^{2}, y = a b$ then the value of $a^{2} - b^{2}$ is

\sqrt{x^{2} + 2 y^{2}}

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\sqrt{x^{2} + 4 y^{2}}

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\sqrt{x^{2} - 4 y^{2}}

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\sqrt{x^{2} - 2 y^{2}}

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Solution

The correct option is C $\sqrt{x^{2} - 4 y^{2}}$
Given $x = a^{2} + b^{2}, y = a b$
Then $a^{2} + b^{2} + 2 a b = x + 2 y \Rightarrow (a + b)^{2} x + 2 y \Rightarrow a + b = \sqrt{x + 2 y}$
And $a^{2} + b^{2} - 2 a b = x + 2 y \Rightarrow (a - b)^{2} x + 2 y \Rightarrow a - b = \sqrt{x - 2 y}$
Then $a^{2} - b^{2} = (a + b) (a - b) = \sqrt{x + 2 y} \times \sqrt{x - 2 y} = \sqrt{x^{2} - 4 y^{2}}$

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Similar questions

Which of the following identity can be used to factorize $x^{2} - 4 y^{2}$ ?

\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1, 4 x^{2} + 4 y^{2} = a^{2} (b > a)

(x + 2 y)^{2} = x^{2} + 4 y^{2} + 4 x y

\frac{x^{2}}{(x^{2} + a^{2}) (x^{2} + b^{2})} = k [\frac{a^{2}}{x^{2} + a^{2}} - \frac{b^{2}}{x^{2} + b^{2}}] \Rightarrow k =

(x + 2 y + 3 z)^{2}

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