if x =5-√3/5+√3. and y= 5+√3/5-√3,show that x²-y²=-10√3/11
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We have : x = 5 − 3√5 + 3√ , So
x = 5 − 3√5 + 3√×5 − 3√5 − 3√ = 25 − 53√ − 5 3√ + 325 − 53√ + 5 3√ − 3= 28 − 103√22= 2(14 − 53√)22 = 14 − 53√11
And
y = 5 + 3√5 − 3√ , So
y = 5 + 3√5 − 3√×5 + 3√5 + 3√ = 25 + 53√ + 5 3√ + 325 + 53√ − 5 3√ − 3= 28 + 103√22= 2(14 + 53√)22 = 14 + 53√11
Then,
To show : x2 − y2 = −103√11 , We take L.H.S. :x2 − y2 We know : a2 − b2 =(a+b)(a−b) , So⇒(x+ y)(x− y)Substitute values of 'x' and 'y' and get :⇒(14 − 53√11+14 + 53√11)(14 − 53√11− 14 + 53√11)⇒(14 − 53√ + 14 + 53√11)(14 − 53√ − 14 − 53√11) ⇒(28/11)(− 10√3/ 11)ButL.H.S. ≠R.H.S.
So , Kindly recheck given query and get back to us so that we could help you precisely .
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We have : x = 5 − 3√5 + 3√ , So
x = 5 − 3√5 + 3√×5 − 3√5 − 3√ = 25 − 53√ − 5 3√ + 325 − 53√ + 5 3√ − 3= 28 − 103√22= 2(14 − 53√)22 = 14 − 53√11
And
y = 5 + 3√5 − 3√ , So
y = 5 + 3√5 − 3√×5 + 3√5 + 3√ = 25 + 53√ + 5 3√ + 325 + 53√ − 5 3√ − 3= 28 + 103√22= 2(14 + 53√)22 = 14 + 53√11
Then,
To show : x2 − y2 = −103√11 , We take L.H.S. :x2 − y2 We know : a2 − b2 =(a+b)(a−b) , So⇒(x+ y)(x− y)Substitute values of 'x' and 'y' and get :⇒(14 − 53√11+14 + 53√11)(14 − 53√11− 14 + 53√11)⇒(14 − 53√ + 14 + 53√11)(14 − 53√ − 14 − 53√11) ⇒(28/11)(− 10√3/ 11)ButL.H.S. ≠R.H.S.
So , Kindly recheck given query and get back to us so that we could help you precisely .
Regards
Was this answer helpful
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