How do you solve $- 4 x^{2} + 4000 x = 0$ $-4x^2+4000x=0$ ?

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How do you solve $- 4 x^{2} + 4000 x = 0$ $-4x^2+4000x=0$ ?

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Answer 1 · 2018-03-02T06:24:11

The solutions are $x = 0, 1000$ $x=0,1000$ .

Explanation:

Simplify the quadratic by dividing by $- 4$ $-4$ , then factor out an $x$ $x$ term. Lastly, solve for $x$ $x$ in the expression that equals $0$ $0$ .

$- 4 x^{2} + 4000 x = 0$ $-4x^2+4000x=0$

$\frac{- 4 x^{2} + 4000 x}{- 4} = \frac{0}{- 4}$ $(-4x^2+4000x)/color(red)(-4)=0/color(red)(-4)$

$x^{2} - 1000 x = 0$ $x^2-1000x=0$

$x (x - 1000) = 0$ $x(x-1000)=0$

$x = 0, 1000$ $x=0,1000$

The solutions are $x = 0$ $x=0$ and $x = 1000$ $x=1000$ . You can verify this answer by seeing where the parabola $y = - 4 x^{2} + 4000 x = 0$ $y=-4x^2+4000x=0$ crosses the $x$ $x$ -axis. (It crosses at $0$ $0$ and $1000$ $1000$ .)

graph{-4x^2+4000x [-1000, 2000, -1000000, 1500000]}

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How do you solve $- 4 x^{2} + 4000 x = 0$ $-4x^2+4000x=0$ ?

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How do you solve -4x^2+4000x=0−4x2+4000x=0-4x^2+4000x=0?

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How do you solve $- 4 x^{2} + 4000 x = 0$ $-4x^2+4000x=0$ ?