Question

if a = root a+2b - root a-2b / rooy a +2b -root a-2b , prove that bx²-az+b=0

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Answer 1

$\texttt{\textsf{\large{\underline{Correct Question}:}}}$

If,

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

Prove that,

$\sf \implies b {x}^{2} - ax + b = 0$

$\texttt{\textsf{\large{\underline{Solution}:}}}$

Given:

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

We can also write it as:

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

Applying componendo and dividendo, we get:

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

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

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

Squaring both sides, we get:

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

Again, applying componendo and dividendo, we get:

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

On simplifying, we get:

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

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

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

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

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

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

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

On cross multiplying, we get:

$\sf \implies b({x}^{2} + 1) =ax$

$\sf \implies b{x}^{2} +b=ax$

$\sf \implies b{x}^{2} - ax +b=0$

Hence, Proved.!!

$\texttt{\textsf{\large{\underline{Know More}:}}}$

If a : b and c : d are two ratios such that a : b : : c : d. Then the following results hold true.

1. Invertendo.

$\sf \implies b : a : : d : c$

2. Alternendo.

$\sf \implies a : c : : b : d$

3. Componendo.

$\sf \implies (a + b) : b: : (c + d) : d$

4. Dividendo.

$\sf \implies (a - b) : b: : (c - d) : d$

5. Componendo and dividendo.

$\sf \implies (a + b) : (a - b): : (c + d) :(c - d)$

6. Convertendo.

$\sf \implies a : (a - b): :c:(c - d)$

Answer 2

Step-by-step explanation:

$\bf \underline{Given-}$

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

$\bf \underline{To\: find-}$

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

$\bf \underline{Solution-}$

$\textsf{We have,}$

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

$\textsf{The denominator is : √(a+2b) - √(a-2b)}$

$\textsf{We know that}$

$\textsf{The rationalising factor of : √(p + q) - √(p-q) = √(p+q) + √(p-q).}$

$\textsf{Therefore, the rationalising factor of: √(a+2b) - √(a-2b) = √(a+2b) + √(a-2b).}$

$\textsf{On, rationalising the denominator,we get}$

$\sf{x = \frac{ \sqrt{a + 2b} + \sqrt{a - 2b} }{ \sqrt{a + 2b} - \sqrt{a - 2b} } \times\frac{ \sqrt{a + 2b} + \sqrt{a - 2b} }{ \sqrt{a + 2b} + \sqrt{a - 2b} } }$

$\sf{x = \frac{ (\sqrt{a + 2b} + \sqrt{a - 2b})( \sqrt{a + 2b} + \sqrt{a - 2b}) }{( \sqrt{a + 2b} - \sqrt{a - 2b})( \sqrt{a + 2b} + \sqrt{a - 2b} )} }$

$\sf{x = \frac{ (\sqrt{a + 2b} + \sqrt{a - 2b} {)}^{2} }{( \sqrt{a + 2b} - \sqrt{a - 2b})( \sqrt{a + 2b} + \sqrt{a - 2b} )} }$

$\textsf{★ Now, comparing the denominator with (a-b)(a+b), we get}$

$\sf{ \: \: \: \: \: a = \sqrt{a + 2b} \: and \: b = \sqrt{a - 2b} }$

$\textsf{Using identity (a+b)(a-b) = a²-b², we get}$

$\sf{x = \frac{ (\sqrt{a + 2b} + \sqrt{a - 2b} {)}^{2} }{( \sqrt{a + 2b} {)}^{2} - (\sqrt{a - 2b} {)}^{2} } }$

$\sf{x = \frac{ (\sqrt{a + 2b} + \sqrt{a - 2b} {)}^{2} }{a + 2b - (a + 2b) } }$

$\sf{x = \frac{ a + 2b + a - 2b + 2 \sqrt{ {a}^{2} - {4b}^{2} } }{a + 2b - (a + 2b) } }$

$\Rightarrow\sf{x = \frac{a + \sqrt{ {a}^{2} - {4b}^{2} } }{2b} }$

$\Rightarrow\sf{2bx =a + \sqrt{ {a}^{2} - {4b}^{2} } }$

$\Rightarrow\sf{2bx - a = \sqrt{ {a}^{2} - {4b}^{2} } }$

$\textsf{Squaring on both sides,we get}$

$\sf{(2bx - a {)}^{2} = {a}^{2} - 4 {b}^{2} }$

$\Rightarrow\sf{ {4b}^{2} {x}^{2} + \cancel{{a}^{2}} - 4abx - \cancel{ {a}^{2}} + {4b}^{2} =0}$

$\Rightarrow\sf{ {4b}^{2} {x}^{2} - 4abx + {4b}^{2} =0 \: \: \Rightarrow\sf{4( {b}^{2} {x}^{2} - abx + {4b}^{2} ) =0} }$

$\Rightarrow\sf{{b}^{2} {x}^{2} - abx + {4b}^{2} =0 \: \: \: \: \: \Rightarrow\sf{b( {b}{x}^{2} - ax + {b} ) =0} }$

$\Rightarrow\sf{{b}{x}^{2} - ax + {b} =0}$

$\bf \underline{Hence, proved.}$