if x^2 + 1/x^2 = 7 …
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Payal Singh 8 years, 6 months ago
Given : {tex}x^2+{1\over x^2} = 7{/tex}
To find : {tex}2x^2-{2\over x^2}{/tex}
Solution :
We know,
{tex}(a+b)^2 = a^2+b^2+2ab{/tex} ...(1)
Put a = x and b = {tex}1\over x{/tex}
{tex}(x+{1\over x})^2= x^2+{1\over x^2}+2{/tex}
=> {tex}(x+{1\over x})^2= 7+2=9{/tex}
=> {tex}x+{1\over x}= 3{/tex}...... (2)
Also
{tex}(x-{1\over x})^2= x^2+{1\over x^2}-2{/tex}
{tex}=> (x-{1\over x})^2= 7-2{/tex}= 5
{tex}=> x-{1\over x}= \sqrt 5{/tex} ... (3)
Multiply (2) and (3)
{tex}x^2-{1\over x^2}= 3\sqrt 5{/tex}
Multiply both side by 2
{tex}2x^2-{2\over x^2}= 6\sqrt 5{/tex}
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