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Mathematics
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solve by completing the square
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x^2 +2x -8=0 (x+1)^2 - 1-8=0
x^2 +4x -5=0 (x+2)^2 - 4-5=0
first one as folows: shift the constant to RHS x^2 + 2x = 8 add square of half of the coefficient of x to both sides x^2 + 2x + (1)^2 = 8 + (1)^2 LHS becomes (x + 1)^2 (x + 1)^2 = 9 taking squareroot of both sides x + 1 = root9 x + 1 = +/- 3 shifting 1 to RHS x = -1 +/- 3 so x = -1-3 = -4 and x = -1 +3 = 2
You can follow the same process for the next one.... If u need help tell me....
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okay thank you
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