Question

If (√a+√b) = 17 and (√a-√b)=1,then find √ab​

Answer 1

Answer:

Explain:

Question ⤵

(√a+√b) = 17

$\huge \mathcal \blue{Solution : }$

$\huge \mathbb \purple{To \: be \: known}$

√ab = ?

$\large \orange {\underline{ \bold{Have \: Given:-}}}$

$( \sqrt{a} - \sqrt{b})=1$

$( \sqrt{a}+ \sqrt{b} ) = 17$

$\bold{On \: squaring \: both \: sides: -}$

$( \sqrt{a}+ \sqrt{b} )^{2} = (17)^{2}$

$\large \bold{ \underline{ \green {Formula : - }}}$

$(a + b)^{2} = (a - b)^{2} + 4ab$

$\scriptsize{( \sqrt{a} - \sqrt{ b})^{2} + 4 [(\sqrt{a}) ( \sqrt{b})] = 289}$

$\scriptsize(1)^{2} + 4 \sqrt{ ab} = 289 \: [\because( \sqrt{a} - \sqrt{ b}) = 1]$

$1 + 4 \sqrt{ ab}= 289$

$4 \sqrt{ ab} = 289 - 1$

$4 \sqrt{ ab}= 288$

$\sqrt{ab} = \frac{288}{4}$

$\sqrt{ab} = 72$

I hope it is helpful