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X^2-26x+160=0 completing the square |
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Answer» Answer: look for the coefficient of x^2 and if it is not 1 then divide the equation by the coefficient of x^2 so as to make it 1. next shift the constant to right hand side take the coefficient of x divide it by 2, then square it and add to both sides of the given equation. then solve for x in the given example, a = 1, B = -26 and C = 160 [ a and b are the coefficients of x^2 and x respectively and c is the constant ] x^2 - 26x + (13)^2 = -160 + (13)^2 => x^2 - 26x +169 = -160 + 169 => x^2 - 26x + 169 = 9 => (x - 13)^2 = 9 => x - 13 = √9 => x - 13 = +3 and -3 => x = 3 + 13 = 16 and x - 13 = -3 => x = 10 therefore x = 16 and x = 10 are the solutions |
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