Math, asked by sandy986, 11 months ago

if x is equal to 3 + root 8 and Y is equal to 3 minus root then one upon X square + 1 upon y square is equal to​

Answers

Answered by LovelyG
20

Correct question: If x = 3 + √8 and y = 3 - √8. Find the value of 1/x² + 1/y².

Answer:

\large{\underline{\boxed{\sf \frac{1}{x {}^{2} }  +  \frac{1}{ {y}^{2} }  = 34}}}

Step-by-step explanation:

Given that -

x = 3 + √8

Squaring both the sides-

⇒ x² = (3 + √8)²

⇒ x² = (3)² + 2 * 3 * √8 + (√8)²

⇒ x² = 9 + 6√8 + 8

⇒ x² = 17 + 6√8

And,

y = 3 - √8

Squaring both the sides-

⇒ y² = (3 - √8)²

⇒ y² = (3)² - 2 * 3 * √8 + (√8)²

⇒ y² = 9 - 6√8 + 8

⇒ y² = 17 - 6√8

Now, we have to find:

 \implies \sf  \frac{1}{x {}^{2} }  +  \frac{1}{y {}^{2} }  \\  \\ \implies \sf  \frac{1}{17 + 6 \sqrt{8} }  +  \frac{1}{17 - 6 \sqrt{8} }  \\  \\ \implies \sf  \frac{17 - 6 \sqrt{8} + 17 + 6 \sqrt{8}  }{(17 + 6 \sqrt{8})(17 - 6 \sqrt{8} )}  \\  \\ \implies \sf  \frac{17 + 17}{(17) {}^{2}  - (6 \sqrt{8} ) {}^{2} }  \\  \\ \implies \sf  \frac{34}{289 - 288}  \\  \\ \implies \sf 34

Hence,

\large{\underline{\boxed{\sf \frac{1}{x {}^{2} }  +  \frac{1}{ {y}^{2} }  = 34}}}

Answered by Anonymous
11

Correction:

If x is equal to 3+√8 and this equal to 3-√8,then find 1/x²+1/y²?

Answer:

Given

x = 3+√8

and, y = 3-√8

First,let us find 1/x and 1/y

On rationalising,

 \mathbf{ \frac{1}{x} =  \frac{1}{3 +  \sqrt{8}} =   \frac{1}{3 +  \sqrt{8}}  \times  \frac{3 -  \sqrt{8} }{3 -  \sqrt{8} }  =  \frac{3 -  \sqrt{8} }{3 {}^{2}  -  \sqrt{8 {}^{2} } }  =  \frac{3 -  \sqrt{8} }{9 - 8}  = 3 -  \sqrt{8} }

 \mathbf{ \frac{1}{y}  =  \frac{1}{3 -  \sqrt{8} } =  \frac{1}{3 -  \sqrt{8}}  \times  \frac{3 +  \sqrt{8} }{3 +  \sqrt{8} } =  \frac{3 +  \sqrt{8} }{3 {}^{2}  -  \sqrt{8 {}^{2} } } = 3 +  \sqrt{8}   }

Now,

 \mathbf{ \frac{1}{x {}^{2} } +  \frac{1}{y {}^{2} } } \\   \\ \mathbf{ = (\frac{1}{x} ) {}^{2} + ( \frac{1}{y}) {}^{2}  } \\  \\  = \mathbf{(3 +  \sqrt{8} ) {}^{2}  + (3 -  \sqrt{8} ) {}^{2}}   \\  \\  \mathbf{ = 9 +8 + 6 \sqrt{8}   + 9 + 8 - 6 \sqrt{8} } \\  \\  =  \mathbf{17 + 17} \\  \\   = \mathbf{34}

Thus,

 \mathbf{ \huge \boxed{ \boxed{ \frac{1}{x {}^{2} \: } +  \frac{1}{y {}^{2} }  = 34 }}}

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