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

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

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Answer 1 · 2015-04-03T09:49:30

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 $x^{2} - 8 x$ $x^2-8x$ , this means that $2 a x = - 8 x$ $2ax=-8x$ , and so $a = - 4$ $a=-4$ .
Adding $19$ $19$ at both sides, we have
$x^{2} - 8 x + 16 = 19$ $x^2-8x+16=19$
Which is the form we wanted, because now we have
${(x - 4)}^{2} = 19$ $(x-4)^2=19$
Which leads us to
$x - 4 = \pm \sqrt{19}$ $x-4=\pm\sqrt{19}$ and finally $x = \pm \sqrt{19} + 4$ $x=\pm\sqrt{19}+4$

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

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