Question

pls answer the question in the attachment stepwise

Answer 1

${\large\bf{\mid{\overline{\underline{Given:-}}}\mid}}$

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

$\Large\underline{\underline{\sf{To\:Prove}:}}$

• 3bx² - 4ax + 3b = 0

$\huge{\mathfrak{\overbrace{\underbrace{\pink{\fbox{\green{\blue{\bf\:S}\pink{o}\red{l}\orange{u}\purple{t}\blue{i}\green{o}\red{n}}}}}}}}$

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

Using componendo and dividendo , we get ,

$\purple\longmapsto\tt \:\dfrac{x + 1}{x - 1}=\dfrac{\sqrt{2a + 3b}}{ \sqrt{2a - 3b}} \\\\\red{\bf\:Squaring\:Both\:Sides}, \\ \\\purple\longmapsto\tt \: \frac{(x + 1)^{2}}{(x - 1)^{2}} = \frac{2a + 3b}{2a - 3b}\\\\\purple\longmapsto\tt \: \frac{(x^{2}+2x +1)}{(x^{2}-2x + 1)}=\frac{2a + 3b}{2a - 3b}\\\\ \green{\bf\:cross \: multiply} \\ \\ \purple\longmapsto\tt \:(2a-3b)(x^{2}+2x +1) = (2a+3b)(x^{2} -2x +1)$

$\purple\longmapsto\tt\:(\cancel{2ax^{2}}+4ax+\cancel2a-3bx^{2}- \cancel{6bx}-3b)=( \cancel{2ax^{2}}-4ax+\cancel2a + 3bx^{2}- \cancel{6bx}+3b) \\\\\purple\longmapsto\tt\:-3bx^{2}+ 4ax-3b=3bx^{2}-4ax+3b\\\\\purple\longmapsto\tt \: 3bx^{2}+3bx^{2} - 4ax - 4ax + 3b + 3b = 0\\\\\purple\longmapsto\tt\:6bx^{2}- 8ax+6b=0\\\\\purple\longmapsto\tt2(3bx^{2}-4ax+3b)=0\\\\\purple\longmapsto \red{\boxed{\tt\:(3bx^{2}-4ax+3b)=0}}$

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$\bold{\boxed{\huge{\boxed{\orange{\small{\boxed{\huge{\red{\bold{\huge\bold{\red{\ddot{\smile}}}}}}}}}}}}}$

Answer 2

$\large \underline{ \blue{ \boxed{ \bf \green{Required \: Answer}}}}$

STEPWISE ANSWER IN ATTACHMENT