Question

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

Prove that: $\large\boxed{bx^2-3ax+b=0}$

Answer 1

Answer:

$\huge\boxed{Hey\:mate\:here\:is\:your\:answer :}$

$\large\red{Solution:}$

$\large\boxed{Given:}$

→$\large\boxed{\frac{x}{1}=\frac{\sqrt{3a+2b}+\sqrt{3a-2b} }{\sqrt{3a+2b}-\sqrt{3a-2b} },}$

→$\large\boxed{\frac{x+1}{x-1}=\frac{\sqrt{3a+2b}+\sqrt{3a-2b}+\sqrt{3a+2b}-\sqrt{3a-2b} }{\sqrt{3a+2b}+\sqrt{3a-2b}-\sqrt{3a+2b}+\sqrt{3a-2b} },}$

[Applying componendo and dividendo],

→$\large\boxed{\frac{x+1}{x-1}=\frac{2\sqrt{3a+2b} }{2\sqrt{3a-2b} },}$

→$\large\boxed{\frac{x+1}{x-1}=\frac{\sqrt{3a+2b} }{3a-2b},}$

→$\large\boxed{\frac{x^2+2x+1}{x^2-2x+1} = \frac{3a+2b}{3a-2b},}$         [Squaring both the sides]

→$\large\boxed{\frac{x^2+2x+1+x^2-2x+1}{x^2+2x+1-x^2+2x-1}=\frac{3a+2b+3a-2b}{3a+2b-3a+2b}}$,

[By componendo and dividendo]

→$\large\boxed{\frac{2(x^2+1)}{4x}=\frac{6a}{4b},}$,

→$\large\boxed{\frac{x^2+1}{2x}=\frac{3a}{2b}}$,

→∴$\large\boxed{2bx^2+2b=9ax}$,

→$\huge\red{bx^2-3ax+b=0}$.

$\huge\boxed{Hope\:it\:helps\:you}$

Answer 2

Here is Your answer...❤

ANSWER

x=

a+2b − a−2b

a+2b +a−2b

= a+2b − a−2b

a+2b + a−2b

× a+2b

+ a−2b

a+2b + a−2b

= (a+2b)−(a−2b)

( a+2b + a−2b )2

= 2ba+ a 2 −4b

⇒2bx=a+ a 2 −4b 2

⇒2bx−a= a 2 −4b 2

Squaring both sides, we get

(2bx−a) 2 =a 2 −4b 2

⇒4b 2 x 2 +a 2 −4abx=a

⇒4b 2 x 2

−4abx+4b

2 =0

⇒bx 2

−ax+b=0

Hope it helps....

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