Question

if y=3-2 root 2 , find y^2+1/y^2

Answer 1

Answer:

$\frac{ {y}^{2} + 1}{ {y}^{2} } = \frac{2}{1} = 2$✔✔

Step-by-step explanation:

$y = 3 - 2\sqrt{2} \: \: \: ( given) \\ \frac{ {y}^{2} + 1}{ {y}^{2} } \\ = \frac{ ({3 - 2\sqrt{2} })^{2} + 1 }{ ({3 - 2\sqrt{2} })^{2} } \\ = \frac{9 - 8 + 1}{9 - 8} \\ = \frac{1 + 1}{1} \\ = \frac{2}{1} \\ = 2$

Answer 2

Solution :-

As given :-

y = 3 - 2√2

We have to find out the value of y² + 1/y²

Now via Rationalising the denominator of the 1/y

$\dfrac{1}{y} = \dfrac{1}{3 - 2\sqrt{2}} \times \dfrac{3 + 2\sqrt{2}}{3 + 2\sqrt{2}}$

$\dfrac{1}{y} = \dfrac{3 + 2\sqrt{2}}{3^2 - (2\sqrt{2})^2}$

$\dfrac{1}{y} = \dfrac{3 + 2\sqrt{2}}{9 - 8}$

$\dfrac{1}{y} = 3 + 2\sqrt{2}$

Now as we know that

a² + b² = (a + b)² - 2ab

$y^2 + \dfrac{1}{y^2} = \left( y + \dfrac{1}{y} \right)^2 - 2 \: . \: y \: . \dfrac{1}{y}$

$y^2 + \dfrac{1}{y^2} = ( 3 - 2\sqrt{2} + 3 + 2\sqrt{2} )^2 - 2$

$y^2 + \dfrac{1}{y^2} = ( 6 )^2 - 2$

$y^2 + \dfrac{1}{y^2} = 36 - 2$

So

$\huge{\boxed{\sf{ y^2 + \dfrac{1}{y^2} = 34}}}$