Mathematics
Neveah21
7

x^2-14x-4=0, solve by completing the square

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(2) Answers
camzastricko

×^2-14×+49-49-4=0 (×-7)^2-53=0 (×-7-7.3)(×-7+7.3) (×-14.3)(×+0.3)=0

ItssHannahh

The answer is: x = 7 - √53      or      x = 7 + √53 The general quadratic equation is: ax² + bx + c  = 0. But, by completing the square we turn it into: a(x + d)² + e = 0, where: d = b/2a e = c - b²/4a Our quadratic equation is x² - 14x -4 = 0, which is after rearrangement: So, a = 1, b = -14, c = -4 Let's first calculate d and e: d = b/2a = -14/2*1 = -14/2 = -7 e = c - b²/4a = -4 - (-14)²/4*1 = -4 - 196/4 = -4 - 49 = -53 By completing the square we have: a(x + d)² + e = 0 1(x + (-7))² + (-53) = 0 (x - 7)² - 53 = 0 (x - 7)² = 53 x - 7 = +/-√53 x = 7 +/- √53 Therefore, the solutions are: x = 7 - √53 or x = 7 + √53

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