Question

If x=√3a+2b + √3a-2b upon √3a+2b - √3a-2b prove that : bx²-3ax+b=0​

Answer 1

$\textbf{Given:}$

$\mathrm{x=\dfrac{\sqrt{3a+2b}+\sqrt{3a-2b}}{\sqrt{3a+2b}-\sqrt{3a-2b}}}$

$\textbf{To prove:}$

$\mathrm{bx^2-3ax+b=0}$

$\textbf{Solution:}$

$\text{Consider,}$

$\mathrm{x=\dfrac{\sqrt{3a+2b}+\sqrt{3a-2b}}{\sqrt{3a+2b}-\sqrt{3a-2b}}}$

$\text{To rationalize the denominator}$

$\mathrm{x=\dfrac{\sqrt{3a+2b}+\sqrt{3a-2b}}{\sqrt{3a+2b}-\sqrt{3a-2b}}{\times}\dfrac{\sqrt{3a+2b}+\sqrt{3a-2b}}{\sqrt{3a+2b}+\sqrt{3a-2b}}}$

$\mathrm{x=\dfrac{(\sqrt{3a+2b}+\sqrt{3a-2b})^2}{(3a+2b)-(3a-2b)}}$

$\mathrm{x=\dfrac{(\sqrt{3a+2b}+\sqrt{3a-2b})^2}{4b}}$

$\mathrm{4b\,x=(\sqrt{3a+2b}+\sqrt{3a-2b})^2}$

$\mathrm{4b\,x=(3a+2b)+(3a-2b)+2\,\sqrt{(3a+2b)(3a-2b)}}$

$\mathrm{4b\,x=6a+2\,\sqrt{9a^2-4b^2}}$

$\mathrm{4b\,x-6a=2\,\sqrt{9a^2-4b^2}}$

$\mathrm{2b\,x-3a=\sqrt{9a^2-4b^2}}$

$\text{Squaring on bothsides, we get}$

$\mathrm{(2b\,x-3a)^2=9a^2-4b^2}$

$\mathrm{4b^2x^2+9a^2-12ab\,x=9a^2-4b^2}$

$\mathrm{4b^2x^2-12ab\,x+4b^2=0}$

$\text{Divide bothsides by 4b}$

$\implies\boxed{\mathrm{\bf\,bx^2-3a\,x+b=0}}$

Answer 2

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

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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