How do you use the quadratic formula to solve $x^{2} + 2 x = 4 x$ $x^2+2x=4x$ ?

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How do you use the quadratic formula to solve $x^{2} + 2 x = 4 x$ $x^2+2x=4x$ ?

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Answer 1 · 2017-05-07T23:30:51

For $a x^{2} + b x + c = 0$ $ax^2+bx+c=0$ , the quadratic formula is $x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $x=(-b+-sqrt(b^2-4ac))/(2a)$ .

Explanation:

First we need to get the equation to standard quadratic form by subtracting 4x from both sides:
$x^{2} - 4 x + 2 x = 0$ $x^2-4x+2x=0$ which we can simplify by combining like-terms
$x^{2} - 2 x = 0$ $x^2-2x=0$
Here we can see that according to the standard quadratic form, a=1, b=-2, and c=0.
By plugging these values into the quadratic formula, $x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $x=(-b+-sqrt(b^2-4ac))/(2a)$
$x = \frac{- (- 2) \pm \sqrt{{(- 2)}^{2} - 4 \cdot 1 \cdot 0}}{2 \cdot 1}$ $x=(-(-2)+-sqrt((-2)^2-4*1*0))/(2*1)$
$x = \frac{2 \pm \sqrt{4}}{2}$ $x=(2+-sqrt(4))/2$
$x = \frac{2 \pm 2}{2}$ $x=(2+-2)/2$
$x = \frac{1 \pm 1}{1}$ $x=(1+-1)/1$
so $x = 0$ $x=0$ or $x = 2$ $x=2$

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How do you use the quadratic formula to solve $x^{2} + 2 x = 4 x$ $x^2+2x=4x$ ?

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