Question

can u do this question ​

Answer 1

$\star$ Question : If   $P=3-2\sqrt{2}$ . Then value of  $P^2+\frac{1}{P^2} =?$

$\star$ Given : $P=3-2\sqrt{2}$

$\star$ To Find :  $P^2+\frac{1}{P^2} =?$

$\star$ Solution :

$\star$ Here ,  $P=3-2\sqrt{2}$

$\star$ Now ,   $\frac{1}{P}=\frac{1}{3-2\sqrt{2} }$

$\star$ Now , Rationalizing the denominator , we get ,

   $=>\frac{1}{P}=\frac{1}{3-2\sqrt{2} }*\frac{3+2\sqrt{2}}{3+2\sqrt{2}} [Since,(a+b)(a-b)=a^2-b^2] \\\\\ =>\frac{1}{P}=\frac{3+2\sqrt{2}}{9-8}\\\\ =>\frac{1}{P}=\frac{3+2\sqrt{2}}{1}\\\\=>\frac{1}{P}=3+2\sqrt{2}$

   $Now,P+\frac{1}{P} =(3-2\sqrt{2}+3+2\sqrt{2})\\\\=>P+\frac{1}{P}=6$

$\star$ Now , squaring on both sides , we get ,

    $=>(P+\frac{1}{P})^2=6^2\\\\=>P^2+\frac{1}{P^2}+2.P.\frac{1}{P}=36\\\\ =>P^2+\frac{1}{P^2}+2=36\\\\ =>P^2+\frac{1}{P^2}=36-2\\\\=>P^2+\frac{1}{P^2}=34$

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