Here -x- in the following is - -a-x-a-x-a-x-a-x-b

The correct option is B 2ab(b^2+1) √(a+x)+√(a x)√(a+x) √(a x)=b1 ⇒√(a+x)√(a x)=b+1b 1 ⇒a+xa x=(b+1)^2(b 1)^2 ⇒(a+x)+(a x)(a+x) (a x)=( ...

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Byju's Answer

Standard X

Mathematics

Componendo and Dividendo

Here 'x' in t...

Question

Here 'x' in the following is : $\frac{\sqrt{a + x} + \sqrt{a - x}}{\sqrt{a + x} - \sqrt{a - x}} = b$

\frac{2 a b}{(b^{2} + 1)}

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\frac{2 a b}{a + b}

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\frac{a + b}{2 a b}

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\frac{b^{2} + 1}{2 a b}

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Solution

The correct option is B $\frac{2 a b}{(b^{2} + 1)}$
$\frac{\sqrt{a + x} + \sqrt{a - x}}{\sqrt{a + x} - \sqrt{a - x}} = \frac{b}{1}$
$\Rightarrow \frac{\sqrt{a + x}}{\sqrt{a - x}} = \frac{b + 1}{b - 1}$
$\Rightarrow \frac{a + x}{a - x} = \frac{(b + 1)^{2}}{(b - 1)^{2}}$
$\Rightarrow \frac{(a + x) + (a - x)}{(a + x) - (a - x)} = \frac{(b + 1)^{2} + (b - 1)^{2}}{(b + 1)^{2} - (b - 1)^{2}}$

$\Rightarrow \frac{2 a}{2 x} = \frac{2 (b^{2} + 1)}{4 b}$
$\Rightarrow x = \frac{2 a b}{b^{2} + 1}$

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Similar questions

Q. Complete the statements by filling in the blanks.

(1) (a + b)² = ........ + 2ab + b²
(2) (a – b)² = a² – 2ab + ..........
(3) (......... + .........) (a – b) = a² – b²
(4) (x + 5)² = ......... + ......... + ..........

If A = $[\begin{matrix} a & b b & a \end{matrix}]$ and $A^{2}$ = $[\begin{matrix} x & y y & x \end{matrix}]$ , then

Which of the following is correct?
a) $(a - b)^{2} = a^{2} + 2 a b - b^{2}$
b) $(a - b)^{2} = a^{2} - 2 a b + b^{2}$
c) $(a - b)^{2} = a^{2} - b^{2}$
d) $(a + b)^{2} = a^{2} + 2 a b - b^{2}$

Q. Show that

\frac{x^{2} + 2 x y + y^{2} - a^{2} + 2 a b - b^{2}}{(x + y - a + b)} = (x + y + a - b)

Simplify $\frac{a - b}{a} + \frac{a + b}{b}$

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Here 'x' in the following is : √a+x+√a−x√a+x−√a−x=b

The correct option is B 2ab(b2+1)√a+x+√a−x√a+x−√a−x=b1⇒√a+x√a−x=b+1b−1⇒a+xa−x=(b+1)2(b−1)2⇒(a+x)+(a−x)(a+x)−(a−x)=(b+1)2+(b−1)2(b+1)2−(b−1)2⇒2a2x=2(b2+1)4b⇒x=2abb2+1

Here 'x' in the following is : $\frac{\sqrt{a + x} + \sqrt{a - x}}{\sqrt{a + x} - \sqrt{a - x}} = b$