How do you solve $4 x^{2} + 40 x + 280 = 0$ $4x^2+40x+280=0$ by completing the square?

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How do you solve $4 x^{2} + 40 x + 280 = 0$ $4x^2+40x+280=0$ by completing the square?

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Answer 1 · 2017-07-02T02:21:58

You can't, unless you want to have complex numbers. If you want complex numbers, then the answer would be $x = - 5 \pm 3 i \sqrt{5}$ $x=-5\pm3i\sqrt{5}$

Explanation:

You can simplify the quadratic to make completing the square easier:
$4 x^{2} + 40 x + 280 = 0$ $4x^2+40x+280=0$
$x^{2} + 10 x + 70 = 0$ $x^2+10x+70=0$

Completing the square:
$x^{2} + 10 x + 70 = 0$ $x^2+10x+70=0$
$x^{2} + 10 x = - 70$ $x^2+10x=-70$
$x^{2} + 10 x + {(\frac{10}{2})}^{2} = - 70 + {(\frac{10}{2})}^{2}$ $x^2+10x+(10/2)^2=-70+(10/2)^2$
$x^{2} + 10 x + 25 = - 70 + 25$ $x^2+10x+25=-70+25$
${(x + 5)}^{2} = - 45$ $(x+5)^2=-45$

We cannot continue here, since there is no way for the above equation to be true.

However, if you want to have complex numbers, we can continue:
$x + 5 = \pm \sqrt{- 45}$ $x+5=\pm\sqrt{-45}$
$x + 5 = \pm i \sqrt{45}$ $x+5=\pmi\sqrt{45}$
$x + 5 = \pm 3 i \sqrt{5}$ $x+5=\pm3i\sqrt{5}$
$x = - 5 \pm 3 i \sqrt{5}$ $x=-5\pm3i\sqrt{5}$

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How do you solve $4 x^{2} + 40 x + 280 = 0$ $4x^2+40x+280=0$ by completing the square?

1 Answer

Explanation:

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How do you solve $4 x^{2} + 40 x + 280 = 0$ $4x^2+40x+280=0$ by completing the square?