Question

if 3bx^2-4ax+3b=0,than prove x=√2a+3b+√2a+3b/√2a+3b-√2a-3b

Answer 1

Answer:

$\huge\colorbox{yellow}{Correct\:Question\:-}$

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

$Then\: \: prove \: \:that \: {3bx}^{2} - 4ax + 3b = 0$

$\huge\colorbox{yellow}{Solution\:-}$

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

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

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

$\frac{ {x}^{2} + 2x + 1 }{ {x}^{2} - 2x + 1 } = \frac{2a + 3b}{2a - 3b}$

${ - 3bx}^{2} - 3b + 4ax = - 4ax + {3bx}^{2} + 3b$

${6bx}^{2} - 8ax + 6b = 0$

$3( {3bx}^{2} - 4ax + 3b) = 0$

$= > {3bx}^{2} - 4ax + 3b = 0$

$\boxed{\pink{ = > {3bx}^{2} - 4ax + 3b = 0}}$

$\huge\colorbox{yellow}{Thank\:You}$