Math, asked by darkghost72, 1 year ago

if x=1/3-√8,find the value of x^3-2x^2-7x+5

Answers

Answered by Anonymous
1
x³-2x²-7x+5
=(1/3-√8)³-2(1/3-√8)²-7(1/3-√8)+5
(a-b)³=a³-3ab(a-b)-b³ 
(a-b)1=a²-2ab+b²
=(1/3)³-3(1/3)(√8)(1/3-√8)-(√8)³-2[(1/3)²-2(1/3)(√8)+(√8)²]-(7/3-7√8)+5
=(1/27)-(√8/3)-8-8√8-2/9+(4/3)√8-8-7/3+7√8+5
=(-364/27)
Answered by Shreyanshijaiswal81
2

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} )

x = 3 +  \sqrt{8}  =>x - 3 =  \sqrt{8} \\  =>(x - 3)^{2}  =  \sqrt{( {8})^{2} } = 8 \\  => {x}^{2}  + 9 - 6x = 8=> {x}^{2}  - 6x + 1 = 0

 {x}^{3}  -  {2x}^{2}  - 7x + 5 = x( {x}^{2}  - 6x + 1) + 4( {x}^{2} 6x + 1) + 16x + 1 \\  = x \times 0 \times  + 4 \times 0 + 16(3 +  \sqrt{8} ) + 1 \\ =  48 + 16 \sqrt{8}  + 1 = 49 + 32 \sqrt{2}

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