Question

If x=√3a+2b+√3a-2b / √3a+2b-√3a-2b , then show that bx^2-ax+b=0

Answer 1

SOLUTION

CORRECT QUESTION

$\displaystyle \sf{If \: \: x = \frac{ \sqrt{3a + 2b} + \sqrt{3a - 2b} }{\sqrt{3a + 2b} - \sqrt{3a - 2b}} }$

then show that

$\sf{ b {x}^{2} - 3ax + b = 0\: }$

GIVEN

$\displaystyle \sf{ x = \frac{ \sqrt{3a + 2b} + \sqrt{3a - 2b} }{\sqrt{3a + 2b} - \sqrt{3a - 2b}} \: \: }$

TO PROVE

$\sf{ b {x}^{2} - 3ax + b = 0\: }$

CALCULATION

It is given that

$\displaystyle \sf{ x = \frac{ \sqrt{3a + 2b} + \sqrt{3a - 2b} }{\sqrt{3a + 2b} - \sqrt{3a - 2b}} \: \: }$

By Componendo - Dividendo Rule

$\displaystyle \sf{ \frac{x + 1}{x - 1} = \frac{ 2\sqrt{3a + 2b} }{2 \sqrt{3a - 2b}} \: \: }$

$\implies \: \displaystyle \sf{ \frac{x + 1}{x - 1} = \frac{ \sqrt{3a + 2b} }{ \sqrt{3a - 2b}} \: \: }$

Squaring both sides

$\displaystyle \sf{ \frac{ {(x + 1)}^{2} }{ {(x - 1)}^{2} } = \frac{ {3a + 2b} }{{3a - 2b}} \: \: }$

$\implies \: \displaystyle \sf{ \frac{ {x}^{2} + 2x + 1}{ {x}^{2} - 2x + 1} = \frac{ {3a + 2b} }{{3a - 2b}} \: \: }$

Again by Componendo Dividendo Rule

$\: \displaystyle \sf{ \frac{ 2( {x}^{2} + 1)}{ 4x } = \frac{ 6a }{4b} \: \: }$

$\implies \: \: \displaystyle \sf{ \frac{ ( {x}^{2} + 1)}{ x } = \frac{ 3a }{b} \: \: }$

$\implies \displaystyle \sf{b {x}^{2} + b = 3ax}$

$\implies \: \sf{ b {x}^{2} - 3ax + b = 0\: }$

Hence the prove follows

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