Solve for p (x-p)^2=q

${(x - p)}^{2} = q$

Step 1

Take the square root of both sides of the equation to eliminate the exponent on the left side.

$x - p = \pm \sqrt{q}$

Step 2

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$x - p = \sqrt{q}$

Subtract $x$ from both sides of the equation.

$- p = \sqrt{q} - x$

Divide each term in $- p = \sqrt{q} - x$ by $- 1$ and simplify.

Divide each term in $- p = \sqrt{q} - x$ by $- 1$ .

$\frac{- p}{- 1} = \frac{\sqrt{q}}{- 1} + \frac{- x}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{p}{1} = \frac{\sqrt{q}}{- 1} + \frac{- x}{- 1}$

Divide $p$ by $1$ .

$p = \frac{\sqrt{q}}{- 1} + \frac{- x}{- 1}$

Simplify the right side.

Simplify each term.

Move the negative one from the denominator of $\frac{\sqrt{q}}{- 1}$ .

$p = - 1 \cdot \sqrt{q} + \frac{- x}{- 1}$

Rewrite $- 1 \cdot \sqrt{q}$ as $- \sqrt{q}$ .

$p = - \sqrt{q} + \frac{- x}{- 1}$

Dividing two negative values results in a positive value.

$p = - \sqrt{q} + \frac{x}{1}$

Divide $x$ by $1$ .

$p = - \sqrt{q} + x$

Next, use the negative value of the $\pm$ to find the second solution.

$x - p = - \sqrt{q}$

Subtract $x$ from both sides of the equation.

$- p = - \sqrt{q} - x$

Divide each term in $- p = - \sqrt{q} - x$ by $- 1$ and simplify.

Divide each term in $- p = - \sqrt{q} - x$ by $- 1$ .

$\frac{- p}{- 1} = \frac{- \sqrt{q}}{- 1} + \frac{- x}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{p}{1} = \frac{- \sqrt{q}}{- 1} + \frac{- x}{- 1}$

Divide $p$ by $1$ .

$p = \frac{- \sqrt{q}}{- 1} + \frac{- x}{- 1}$

Simplify the right side.

Simplify each term.

Dividing two negative values results in a positive value.

$p = \frac{\sqrt{q}}{1} + \frac{- x}{- 1}$

Divide $\sqrt{q}$ by $1$ .

$p = \sqrt{q} + \frac{- x}{- 1}$

Dividing two negative values results in a positive value.

$p = \sqrt{q} + \frac{x}{1}$

Divide $x$ by $1$ .

$p = \sqrt{q} + x$

The complete solution is the result of both the positive and negative portions of the solution.

$p = - \sqrt{q} + x$

$p = \sqrt{q} + x$

$p = - \sqrt{q} + x$

$p = \sqrt{q} + x$

Solve for p (x-p)^2=q

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