Find All Complex Solutions square root of 9x+10=x

$\sqrt{9 x + 10} = x$

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

${\sqrt{9 x + 10}}^{2} = x^{2}$

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{9 x + 10}$ as ${(9 x + 10)}^{\frac{1}{2}}$ .

${({(9 x + 10)}^{\frac{1}{2}})}^{2} = x^{2}$

Simplify the left side.

Simplify ${({(9 x + 10)}^{\frac{1}{2}})}^{2}$ .

Multiply the exponents in ${({(9 x + 10)}^{\frac{1}{2}})}^{2}$ .

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

${(9 x + 10)}^{\frac{1}{2} \cdot 2} = x^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(9 x + 10)}^{\frac{1}{2} \cdot 2} = x^{2}$

Rewrite the expression.

${(9 x + 10)}^{1} = x^{2}$

Simplify.

$9 x + 10 = x^{2}$

Solve for $x$ .

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

$9 x + 10 - x^{2} = 0$

Factor the left side of the equation.

Factor $- 1$ out of $9 x + 10 - x^{2}$ .

Reorder the expression.

Move $10$ .

$9 x - x^{2} + 10 = 0$

Reorder $9 x$ and $- x^{2}$ .

$- x^{2} + 9 x + 10 = 0$

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

$- (x^{2}) + 9 x + 10 = 0$

Factor $- 1$ out of $9 x$ .

$- (x^{2}) - (- 9 x) + 10 = 0$

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

$- (x^{2}) - (- 9 x) - 1 \cdot - 10 = 0$

Factor $- 1$ out of $- (x^{2}) - (- 9 x)$ .

$- (x^{2} - 9 x) - 1 \cdot - 10 = 0$

Factor $- 1$ out of $- (x^{2} - 9 x) - 1 (- 10)$ .

$- (x^{2} - 9 x - 10) = 0$

Factor.

Factor $x^{2} - 9 x - 10$ 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 $- 9$ .

$- 10, 1$

Write the factored form using these integers.

$- ((x - 10) (x + 1)) = 0$

Remove unnecessary parentheses.

$- (x - 10) (x + 1) = 0$

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

$x - 10 = 0$

$x + 1 = 0$

Set $x - 10$ equal to $0$ and solve for $x$ .

Set $x - 10$ equal to $0$ .

$x - 10 = 0$

Add $10$ to both sides of the equation.

$x = 10$

Set $x + 1$ equal to $0$ and solve for $x$ .

Set $x + 1$ equal to $0$ .

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

The final solution is all the values that make $- (x - 10) (x + 1) = 0$ true.

$x = 10, - 1$

Exclude the solutions that do not make $\sqrt{9 x + 10} = x$ true.

$x = 10$

Find All Complex Solutions square root of 9x+10=x

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