How do you solve $x^{2} + 9 x + 20.25 = 0$ $x^2+9x+20.25=0$ by completing the square?

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Answer 1 · 2016-12-18T13:57:11

$x = - \frac{9}{2} = - 4.5$ $x = -9/2 = -4.5$

The left hand side is already a perfect square trinomial:

${(x + \frac{9}{2})}^{2} = x^{2} + 9 x + \frac{81}{4} = x^{2} + 9 x + 20.25$ $(x+9/2)^2 = x^2+9x+81/4 = x^2+9x+20.25$

So the given equation is equivalent to:

${(x + \frac{9}{2})}^{2} = 0$ $(x+9/2)^2 = 0$

Hence:

$x + \frac{9}{2} = 0$ $x+9/2 = 0$

Hence:

$x = - \frac{9}{2}$ $x = -9/2$

How do you solve x^2+9x+20.25=0x2+9x+20.25=0x^2+9x+20.25=0 by completing the square?