Solve for p square root of -10+7p=p

$\sqrt{- 10 + 7 p} = p$

Step 1

To remove the radical on the left side of the equation, square both sides of the equation.

${\sqrt{- 10 + 7 p}}^{2} = p^{2}$

Step 2

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{- 10 + 7 p}$ as ${(- 10 + 7 p)}^{\frac{1}{2}}$ .

${({(- 10 + 7 p)}^{\frac{1}{2}})}^{2} = p^{2}$

Simplify the left side.

Simplify ${({(- 10 + 7 p)}^{\frac{1}{2}})}^{2}$ .

Multiply the exponents in ${({(- 10 + 7 p)}^{\frac{1}{2}})}^{2}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

${(- 10 + 7 p)}^{\frac{1}{2} \cdot 2} = p^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(- 10 + 7 p)}^{\frac{1}{2} \cdot 2} = p^{2}$

Rewrite the expression.

${(- 10 + 7 p)}^{1} = p^{2}$

Simplify.

$- 10 + 7 p = p^{2}$

Step 3

Solve for $p$ .

Subtract $p^{2}$ from both sides of the equation.

$- 10 + 7 p - p^{2} = 0$

Factor the left side of the equation.

Factor $- 1$ out of $- 10 + 7 p - p^{2}$ .

Reorder the expression.

Move $- 10$ .

$7 p - p^{2} - 10 = 0$

Reorder $7 p$ and $- p^{2}$ .

$- p^{2} + 7 p - 10 = 0$

Factor $- 1$ out of $- p^{2}$ .

$- (p^{2}) + 7 p - 10 = 0$

Factor $- 1$ out of $7 p$ .

$- (p^{2}) - (- 7 p) - 10 = 0$

Rewrite $- 10$ as $- 1 (10)$ .

$- (p^{2}) - (- 7 p) - 1 \cdot 10 = 0$

Factor $- 1$ out of $- (p^{2}) - (- 7 p)$ .

$- (p^{2} - 7 p) - 1 \cdot 10 = 0$

Factor $- 1$ out of $- (p^{2} - 7 p) - 1 (10)$ .

$- (p^{2} - 7 p + 10) = 0$

Factor.

Factor $p^{2} - 7 p + 10$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $10$ and whose sum is $- 7$ .

$- 5, - 2$

Write the factored form using these integers.

$- ((p - 5) (p - 2)) = 0$

Remove unnecessary parentheses.

$- (p - 5) (p - 2) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$p - 5 = 0$

$p - 2 = 0$

Set $p - 5$ equal to $0$ and solve for $p$ .

Set $p - 5$ equal to $0$ .

$p - 5 = 0$

Add $5$ to both sides of the equation.

$p = 5$

Set $p - 2$ equal to $0$ and solve for $p$ .

Set $p - 2$ equal to $0$ .

$p - 2 = 0$

Add $2$ to both sides of the equation.

$p = 2$

The final solution is all the values that make $- (p - 5) (p - 2) = 0$ true.

$p = 5, 2$

Solve for p square root of -10+7p=p

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