How do you solve by completing the square $x^{2} + 3 x - 8 = 0$ $x^2+3x-8=0$ ?

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How do you solve by completing the square $x^{2} + 3 x - 8 = 0$ $x^2+3x-8=0$ ?

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Answer 1 · 2015-04-03T22:02:17

Solving a quadratic expression by completing the square means to manipulate the expression in order to write it in the form
${(x + a)}^{2} = b$ $(x+a)^2=b$
So, if $b \geq 0$ $b\ge 0$ , you can take the square root at both sides to get
$x + a = \pm \sqrt{b}$ $x+a=\pm\sqrt{b}$
and conclude $x = \pm \sqrt{b} - a$ $x=\pm\sqrt{b}-a$ .

Now, we have ${(x + a)}^{2} = x^{2} + 2 a x + a^{2}$ $(x+a)^2=x^2+2ax+a^2$ . Since you equation starts with $x^{2} + 3 x$ $x^2+3x$ , this means that $2 a x = 3 x$ $2ax=3x$ , and so $a = \frac{3}{2}$ $a=3/2$ .
Adding $\frac{41}{4}$ $41/4$ at both sides, we have
$x^{2} + 3 x + \frac{9}{4} = \frac{41}{4}$ $x^2+3x+9/4=41/4$
Which is the form we wanted, because now we have
${(x + \frac{3}{2})}^{2} = \frac{41}{4}$ $(x+3/2)^2=41/4$
Which leads us to
$x + \frac{3}{2} = \pm \sqrt{\frac{41}{4}} = \pm \frac{\sqrt{41}}{2}$ $x+3/2=\pm\sqrt{41/4}=\pm \sqrt{41}/2$ and finally $x = \pm \frac{\sqrt{41}}{2} - \frac{3}{2}$ $x=\pm\sqrt{41}/2-3/2$

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How do you solve by completing the square $x^{2} + 3 x - 8 = 0$ $x^2+3x-8=0$ ?

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