Question

if x equal to 3 plus 2 under root 2 then find x raised to the power 2 plus 1 by x raised to the power 2​

Answer 1

Answer:

➣ Given

x = 3+2$\sqrt{2}$

➣ To find

x² + $\dfrac{1}{x²}$

➣ By using identity

$\small\boxed{(a + b)² = a²+ b²+ 2ab}$

☞ put the value of x

⟹ x² + $\dfrac{1}{x²}$ + 2 (x) * $\dfrac{1}{x}$

⟹. (3 + 2√2)² + $\dfrac{1}{(3+2√2)²}$ + 2

⟹ 9 + 8 + $\dfrac{1}{9 + 8}$ + 2

⟹ 17 +$\dfrac{1}{17}$ + 2

⟹. $\dfrac{290}{17}$ + 2

⟹. $\dfrac{290 + 34}{17}$

⟹. $\dfrac{324}{17}$

Hope it helps

Answer 2

Answer:

$\huge\boxed{\fcolorbox{orange}{blue}{Given:}}$

$x=3+2 \sqrt[]{2}$

$\huge\boxed{\fcolorbox{orange}{blue}{To find:}}$

$x+1/{×}^{2}$

$\huge\boxed{\fcolorbox{orange}{blue}{Solution:}}$

First,find the value of x^2

${x}^{2}=({3+2 \sqrt[]{2}})^{2}\\</p><p>=>{x}^{2}={3}^{2}+{2 \sqrt[]{2}}^{2}+2×3×2 \sqrt[]{2}\\</p><p>=>{x}^{2}=9+8+12 \sqrt[]{2}\\</p><p>Hence,{x}^{2}=17+12 \sqrt[]{2}$

Put the value of {x}^{2}

$And,find\ the\ value\ of\ 1/{x}^{2}=</p><p>1/(17+12 \sqrt[]{2})\\</p><p>(1/(17+12 \sqrt[]{2}))×(17-22 \sqrt[]{2}/17-12 \sqrt[]{2})\\</p><p>through\ rationalising\ denominator\\</p><p>=>(1×17-12 \sqrt[]{2})/({17-12 \sqrt[]{2}}^{2})\\</p><p>=>17-12 \sqrt[]{2}/({17}^{2}-{12 \sqrt[]{2}}^{2}\\</p><p>=>17-12 \sqrt[]{2}/(289-288)\\</p><p>=>17-12 \sqrt[]{2}/1\\</p><p>hence\ the\ value\ of 1/{x}^{2}is 17-12 \sqrt[]{2}$

Now, put the values and solve it,

$x+1/{×}^{2}\\</p><p>=17+12 \sqrt[]{2}+17-12 \sqrt[]{2}\\</p><p>=17+17+12 \sqrt[]{2}-12 \sqrt[]{2}$

=34 is answer...

$\huge\boxed{\fcolorbox{orange}{blue}{Formula used:}}$

${(a+b)}^{2}={a}^{2}+2ab+{b}^{2}\\</h3><h3>{a}^{2}-{b}^{2}=(a+b)(a-b)$