Can you solve for $x$ $x$ in $a x^{2} - b x + c = 0$ $ax^2 - bx + c = 0$ (if a,b,c are constants)?

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Can you solve for $x$ $x$ in $a x^{2} - b x + c = 0$ $ax^2 - bx + c = 0$ (if a,b,c are constants)?

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Answer 1 · 2018-02-16T14:02:08

See a solution process below:

Explanation:

$a x \cdot x - b x + c = 0$ $ax*x - bx + c = 0$ can be rewritten as:

$a x^{2} - b x + c = 0$ $ax^2 - bx + c = 0$

We can use the quadratic equation to solve this problem:

The quadratic formula states:

For $l x^{2} + m x + n = 0$ $color(red)(l)x^2 + color(blue)(m)x + color(green)(n) = 0$ , the values of $x$ $x$ which are the solutions to the equation are given by:

$x = \frac{- m \pm \sqrt{m^{2} - (4 l n)}}{2 \cdot l}$ $x = (-color(blue)(m) +- sqrt(color(blue)(m)^2 - (4color(red)(l)color(green)(n))))/(2 * color(red)(l))$

Substituting:

$a$ $color(red)(a)$ for $l$ $color(red)(l)$

$- b$ $color(blue)(-b)$ for $m$ $color(blue)(m)$

$c$ $color(green)(c)$ for $n$ $color(green)(n)$ gives:

$x = \frac{- - b \pm \sqrt{{- b}^{2} - (4 \cdot a \cdot c)}}{2 \cdot a}$ $x = (-color(blue)(-b) +- sqrt(color(blue)(-b)^2 - (4 * color(red)(a) * color(green)(c))))/(2 * color(red)(a))$

Or

$x = \frac{b \pm \sqrt{{- b}^{2} - (4 \cdot a \cdot c)}}{2 \cdot a}$ $x = (color(blue)(b) +- sqrt(color(blue)(-b)^2 - (4 * color(red)(a) * color(green)(c))))/(2 * color(red)(a))$

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Can you solve for $x$ $x$ in $a x^{2} - b x + c = 0$ $ax^2 - bx + c = 0$ (if a,b,c are constants)?

1 Answer

Explanation:

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Can you solve for $x$ $x$ in $a x^{2} - b x + c = 0$ $ax^2 - bx + c = 0$ (if a,b,c are constants)?