Question

if x4+2x³-4x²-4x+4=0 then 2s1-s2+s3-s4​

Answer 1

Answer:

Step-by-step explanation:

Given : (x + 1) / (x - 1) + (x - 2) / (x + 2) = 4 - (2x + 3) /(x - 2)  

[(x + 1)  (x + 2) + (x - 1) + (x - 2)] / [(x - 1) (x + 2)] =  [4(x - 2) - (2x + 3)] /(x - 2)

[ By taking LCM]

[(x² + 2x + 1x +2) + ( x² - 2x - 1x +2)] / [x² + 2x - 1x -2] = [4x - 8 - 2x - 3] / (x - 2)  

[x² + 3x +2 +x² - 3x +2] / (x² + x - 2) = (2x - 11)/ (x - 2)

(2x² + 4)/(x² + x - 2) = (2x - 11)/ (x - 2)

(2x² + 4)(x - 2) =  (x² + x - 2) (2x - 11)

2x³ - 4x² + 4x - 8 = 2x³ + 2x² - 4x - 11x² - 11x + 22

2x³ - 2x³ - 4x² - 2x² +11x² + 4x + 4x + 11x - 8 - 22 = 0

5x² + 19x - 30 = 0

5x² + 25x - 6x - 30 = 0

5x(x + 25) - 6 (x + 5) = 0

(5x - 6) (x + 5) = 0

(5x - 6)  = 0  (x + 5) = 0

5x = 6  or  x = - 5  

x = 6/5  or  x = - 5  

Hence, the roots of the quadratic equation (x + 1) / (x - 1) + (x - 2) / (x + 2) = 4 - (2x + 3) /(x - 2)  are 6/5  & - 5 .

★★ METHOD TO FIND SOLUTION OF a quadratic equation by FACTORIZATION METHOD :  

We first write the given quadratic polynomial as product of two linear factors by splitting the middle term and then equate each factor to zero to get desired roots of given quadratic equation.

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Answer 2

Answer:

$for \: {x}^{4} + 2 {x}^{3} - 4 {x}^{2} - 4x + 4 = 0$

$sin_1 = -2$

$sin_2 = -4$

$sin_3 = +4$

$sin_4 = +4$

GIVES,

$2s_1 - s_2 + s_3 = - 4 + 4 + 4 - 4 = 0$

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