Question

if y is equals to 3 minus 2 under root 2 then find y square + 1 upon y square​

Answer 1

Answer:

Let us consider \frac{1}{x}=\frac{1}{3-2 \sqrt{2}}x1=3−221

Rationalising the denominator gives  

\begin{gathered}\begin{array}{l}{=\frac{1}{3-2 \sqrt{2}} \times \frac{3+2 \sqrt{2}}{3+2 \sqrt{2}}} \\ {=\frac{3+2 \sqrt{2}}{3^{2}-(2 \sqrt{2})^{2}}} \\ {=\frac{3+2 \sqrt{2}}{9-8}} \\ {=3+2 \sqrt{2}}\end{array}\end{gathered}=3−221×3+223+22=32−(22)23+22=9−83+22=3+22

Now, x^{2}-\frac{1}{x^{2}}=(3-2 \sqrt{2})^{2}-(3+2 \sqrt{2})^{2}x2−x21=(3−22)2−(3+22)2

Applying the algebraic formula (a-b)^2 =a^2+b^2-2ab(a−b)2=a2+b2−2ab

=3^{2}+(2 \sqrt{2})^{2}-2 \times 3 \times 2 \sqrt{2}-\left(3^{2}+(2 \sqrt{2})^{2}+2 \times 3 \times 2 \sqrt{2}\right)=32+(22)2−2×3×22−(32+(22)2+2×3×22)

Simplifying the formula and terms we get

\begin{gathered}\begin{array}{l}{=9+8-12 \sqrt{2}-9-8-12 \sqrt{2} |} \\ {=-24 \sqrt{2}}\end{array}\end{gathered}=9+8−122−9−8−122∣=−242

Therefore, \bold{x^{2}-\frac{1}{x^{2}}=-24 \sqrt{2}}x2−x21=−242  if \bold{x=3-2 \sqrt{2}.}x=3−22.