How do you solve the equation $2 x^{2} + 8 x + 4 = x^{2} + 4 x$ $2x^2+8x+4=x^2+4x$ by completing the square?

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How do you solve the equation $2 x^{2} + 8 x + 4 = x^{2} + 4 x$ $2x^2+8x+4=x^2+4x$ by completing the square?

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Answer 1 · 2017-02-16T20:38:59

The equation has one root $x = - 2$ $x = -2$

Explanation:

First simplify the equation into a standard form
$2 x^{2} + 8 x + 4 = x^{2} + 4 x$ $2x^2+8x+4=x^2+4x$
becomes
$x^{2} + 4 x + 4 = 0$ $x^2 +4x+4 = 0$ (equation 2)

Use the formula
$\frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $(-b+-sqrt(b^2-4ac))/(2a)$

substitute in the values from equation 2
$\frac{- 4 \pm \sqrt{16 - 16}}{2}$ $(-4+-sqrt(16-16))/2$

The equation has one root $x = - 2$ $x = -2$

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How do you solve the equation $2 x^{2} + 8 x + 4 = x^{2} + 4 x$ $2x^2+8x+4=x^2+4x$ by completing the square?

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Explanation:

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How do you solve the equation $2 x^{2} + 8 x + 4 = x^{2} + 4 x$ $2x^2+8x+4=x^2+4x$ by completing the square?