How do you solve $x^{2} + 4 x - 3 = 0$ $x^2+4x-3=0$ by completing the square?

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Answer 1 · 2015-05-11T17:47:05

$0 = x^{2} + 4 x - 3 = x^{2} + 4 x + 4 - 7 = {(x + 2)}^{2} - 7$ $0 = x^2+4x-3 = x^2+4x+4-7 = (x+2)^2-7$

Adding 7 to both ends we get:

${(x + 2)}^{2} = 7$ $(x+2)^2 = 7$

So $x + 2 = \pm \sqrt{7}$ $x + 2 = +-sqrt(7)$

Subtracting 2 from both sides, we arrive at:

$x = \pm \sqrt{7} - 2$ $x = +-sqrt(7)-2$ .

In general, $a x^{2} + b x + c = a {(x + \frac{b}{2 a})}^{2} + (c - \frac{b^{2}}{4 a})$ $ax^2+bx+c = a(x+b/(2a))^2 + (c-b^2/(4a))$ .

How do you solve x2+4x−3=0x^2+4x-3=0 by completing the square?