Question

$\frac{ \sqrt{1 + x} + \sqrt{1 - x} }{ \sqrt{1 + x} - \sqrt{1 - x} } = p \\ proved \: it \: \: \: \: ({p}^{2} - \frac{2p}{x} + 1) = 0$
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Answer 1

Step-by-step explanation:

Given:-

[√(1+x)+√(1-x)]/[√(1+x)-√(1-x)] = p

To find:-

Prove that p^2-(2p/x)+1 = 0

Solution:-

Given that

[√(1+x)+√(1-x)] / [√(1+x)-√(1-x)] = p

The denominator = [√(1+x)-√(1-x)]

Rationalising factor of [√(1+x)-√(1-x)] is

[√(1+x) + √(1-x)]

On Rationalising the denominator then

=> [√(1+x)+√(1-x)] / [√(1+x)-√(1-x)]

× [√(1+x) + √(1-x)] / [√(1+x) + √(1-x)]

=>[√(1+x)+√(1-x)][√(1+x)+√(1-x)] / [√(1+x)-√(1-x)][√(1+x)+√(1-x)]

=> [√(1+x)+√(1-x)]^2/[√(1+x)-√(1-x)]

[√(1+x)+√(1-x)]

=> [√(1+x)+√(1-x)]^2/[{√1+x)}^2+{(1-x)}^2]

Since (a+b)(a-b)=a^2-b^2

=> [√(1+x)+√(1-x)]^2/[(1+x )-(1-x)]

=> [√(1+x)+√(1-x)]^2/(1+x-1+x)

=> [√(1+x)+√(1-x)]^2/2x

We know that (a+b)^2 = a^2+2ab+b^2

=>[ [√(1+x)]^2+2√(1+x)√(1-x)+[√(1-x)]^2]/2x

=> [(1+x)+2√[(1+x)(1-x)]+1-x]/2x

=>[ 2+2√[(1+x)(1-x)]]/2x

=> 2[1+√[(1+x)(1-x)]] /2x

=> [1+√[(1+x)(1-x)]]/x

p = [1+√[(1+x)(1-x)]]/x

On squaring both sides then

=> p^2 = [[1+√[(1+x)(1-x)]]/x]^2

=> p^2 = [1^2+2√[(1+x)(1-x)]+[√[(1+x)(1-x)]]^2] /x^2

=> p^2 =[ 1+(1+x)(1-x)+2√[(1+x)(1-x)]]/x^2

=>p^2 = [2-x^2+2√[(1+x)(1-x)]]/x^2

and

2p/x = 2[1+√[(1+x)(1-x)]]/x×x

=> 2p/x = 2[1+√[(1+x)(1-x)]]/x^2

Now p^2-(2p/x)+1

=> [2-x^2+2√[(1+x)(1-x)]]/x^2-[2[1+√[(1+x)(1-x)]]/x^2]+1

=> [(2-2)/x^2]+(-x^2/x^2) +[2√[(1+x)(1-x)]]/

x^ -2√[(1+x)(1-x)]]/x^2] +1

=> 0+(-1)+0+1

=> -1+1

=>0

p^2-(2p/x)+1 = 0

Hence, Proved

Used formulae:-

• (a+b)^2 = a^2+2ab+b^2
• The product of two irrational numbers is a rational number then each of them are called Rationalising factors.
• The Rationalising factor of√a+√b is √a-√b