Find the Roots (Zeros) square root of 56-x=x

$\sqrt{56 - x} = x$

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

${\sqrt{56 - x}}^{2} = x^{2}$

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{56 - x}$ as ${(56 - x)}^{\frac{1}{2}}$ .

${({(56 - x)}^{\frac{1}{2}})}^{2} = x^{2}$

Simplify the left side.

Simplify ${({(56 - x)}^{\frac{1}{2}})}^{2}$ .

Multiply the exponents in ${({(56 - x)}^{\frac{1}{2}})}^{2}$ .

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

${(56 - x)}^{\frac{1}{2} \cdot 2} = x^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(56 - x)}^{\frac{1}{2} \cdot 2} = x^{2}$

Rewrite the expression.

${(56 - x)}^{1} = x^{2}$

Simplify.

$56 - x = x^{2}$

Solve for $x$ .

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

$56 - x - x^{2} = 0$

Factor the left side of the equation.

Factor $- 1$ out of $56 - x - x^{2}$ .

Reorder the expression.

Move $56$ .

$- x - x^{2} + 56 = 0$

Reorder $- x$ and $- x^{2}$ .

$- x^{2} - x + 56 = 0$

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

$- (x^{2}) - x + 56 = 0$

Factor $- 1$ out of $- x$ .

$- (x^{2}) - (x) + 56 = 0$

Rewrite $56$ as $- 1 (- 56)$ .

$- (x^{2}) - (x) - 1 \cdot - 56 = 0$

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

$- (x^{2} + x) - 1 \cdot - 56 = 0$

Factor $- 1$ out of $- (x^{2} + x) - 1 (- 56)$ .

$- (x^{2} + x - 56) = 0$

Factor.

Factor $x^{2} + x - 56$ 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 $- 56$ and whose sum is $1$ .

$- 7, 8$

Write the factored form using these integers.

$- ((x - 7) (x + 8)) = 0$

Remove unnecessary parentheses.

$- (x - 7) (x + 8) = 0$

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

$x - 7 = 0$

$x + 8 = 0$

Set $x - 7$ equal to $0$ and solve for $x$ .

Set $x - 7$ equal to $0$ .

$x - 7 = 0$

Add $7$ to both sides of the equation.

$x = 7$

Set $x + 8$ equal to $0$ and solve for $x$ .

Set $x + 8$ equal to $0$ .

$x + 8 = 0$

Subtract $8$ from both sides of the equation.

$x = - 8$

The final solution is all the values that make $- (x - 7) (x + 8) = 0$ true.

$x = 7, - 8$

Exclude the solutions that do not make $\sqrt{56 - x} = x$ true.

$x = 7$

Find the Roots (Zeros) square root of 56-x=x

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