Question

If x = √(3a+2b) + √(3a - 2b) / √(3a+2b) - √(3a-2b), prove that : bx² - 3ax + b = 0 ​

Answer 1

$\underline{\Large{ \sf{Given \: information:-}}}$

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

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

• bx² - 3ax + b = 0

$\underline{\Large{ \sf{Let's \: Proof:-}}}$

Taking denominator as 1:

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

Usage of componendo and dividendo:

Which states that,

• $\boxed{\sf{ \dfrac{a \: + b}{a \: - \: b} = \: \dfrac{c \: + \: d}{c \: - \: d} }}$

Using it now,

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

Here we gets,

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

Cancelling 2,

$: \implies \: \sf{ \dfrac{x + 1}{x - 1} \: = \: \dfrac{ \cancel2 \: \sqrt{3a \: + \: 2b} }{ \cancel2 \: \sqrt{3a \: - 2b} } }$

After cancelling 2 we gets,

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

Roots in R.H.S. would be cancelled,

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

Squaring L.H.S. side terms,

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

After squaring we gets,

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

Let us use the componendo and dividendo again,

Concept understanding is written above.

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

On solving them we gets,

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

Here we have even grouped terms.

Dividing both the sides with 2, (i.e., cancelling terms)

$: \implies \: \sf{ \dfrac{ \cancel2 \: (x {}^{2} \: + \: 1) }{ \cancel4x} \: = \: \dfrac{6a}{4b} }$

$: \implies \: \sf{ \dfrac{ \: x {}^{2} \: + \: 1}{ 2x} \: = \: \dfrac{6a}{4b} }$

$: \implies \: \sf{ \dfrac{ \: x {}^{2} \: + \: 1}{ 2x} \: = \: \dfrac{ \cancel6a}{ \cancel4b} }$

We gets now,

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

On cross multiplying we gets,

$: \implies \: \sf{2bx {}^{2} \: + \: 2b \: = \: 6ax }$

On dividing with 2 we gets,

$: \implies \:\sf{bx {}^{2} \: + \: 3ax\: + \: b \: = \: 3ax}$

On transposing sides,

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

Hence proved!!