Solve for x ax^2=bx - MathMaster

$a x^{2} = b x$

Subtract $b x$ from both sides of the equation.

$a x^{2} - b x = 0$

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Substitute the values $a = a$ , $b = - b$ , and $c = 0$ into the quadratic formula and solve for $x$ .

$\frac{b \pm \sqrt{{(- b)}^{2} - 4 \cdot (a \cdot 0)}}{2 a}$

Simplify the numerator.

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Rewrite $4 \cdot (a \cdot 0)$ as ${(2 \cdot 0)}^{2}$ .

$x = \frac{b \pm \sqrt{{(- b)}^{2} - {(2 \cdot 0)}^{2}}}{2 a}$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = - b$ and $b = 2 \cdot 0$ .

$x = \frac{b \pm \sqrt{(- b + 2 \cdot 0) (- b - (2 \cdot 0))}}{2 a}$

Simplify.

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Multiply $2$ by $0$ .

$x = \frac{b \pm \sqrt{(- b + 0) (- b - (2 \cdot 0))}}{2 a}$

Add $- b$ and $0$ .

$x = \frac{b \pm \sqrt{- b (- b - (2 \cdot 0))}}{2 a}$

Combine exponents.

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Multiply $2$ by $0$ .

$x = \frac{b \pm \sqrt{- b (- b - 0)}}{2 a}$

Multiply $- 1$ by $0$ .

$x = \frac{b \pm \sqrt{- b (- b + 0)}}{2 a}$

Add $- b$ and $0$ .

$x = \frac{b \pm \sqrt{- b (- b)}}{2 a}$

Multiply $- 1$ by $- 1$ .

$x = \frac{b \pm \sqrt{1 b \cdot b}}{2 a}$

Multiply $b$ by $1$ .

$x = \frac{b \pm \sqrt{b \cdot b}}{2 a}$

Raise $b$ to the power of $1$ .

$x = \frac{b \pm \sqrt{b \cdot b}}{2 a}$

Raise $b$ to the power of $1$ .

$x = \frac{b \pm \sqrt{b \cdot b}}{2 a}$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$x = \frac{b \pm \sqrt{b^{1 + 1}}}{2 a}$

Add $1$ and $1$ .

$x = \frac{b \pm \sqrt{b^{2}}}{2 a}$

Pull terms out from under the radical, assuming positive real numbers.

$x = \frac{b \pm b}{2 a}$

The final answer is the combination of both solutions.

$x = \frac{b}{a}$

$x = 0$

Solve for x ax^2=bx

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