Solve for x square root of x-5+x=3

$\sqrt{x - 5} + x = 3$

Subtract $x$ from both sides of the equation.

$\sqrt{x - 5} = 3 - x$

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

${\sqrt{x - 5}}^{2} = {(3 - x)}^{2}$

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{x - 5}$ as ${(x - 5)}^{\frac{1}{2}}$ .

${({(x - 5)}^{\frac{1}{2}})}^{2} = {(3 - x)}^{2}$

Simplify the left side.

Simplify ${({(x - 5)}^{\frac{1}{2}})}^{2}$ .

Multiply the exponents in ${({(x - 5)}^{\frac{1}{2}})}^{2}$ .

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

${(x - 5)}^{\frac{1}{2} \cdot 2} = {(3 - x)}^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(x - 5)}^{\frac{1}{2} \cdot 2} = {(3 - x)}^{2}$

Rewrite the expression.

${(x - 5)}^{1} = {(3 - x)}^{2}$

Simplify.

$x - 5 = {(3 - x)}^{2}$

Simplify the right side.

Simplify ${(3 - x)}^{2}$ .

Rewrite ${(3 - x)}^{2}$ as $(3 - x) (3 - x)$ .

$x - 5 = (3 - x) (3 - x)$

Expand $(3 - x) (3 - x)$ using the FOIL Method.

Apply the distributive property.

$x - 5 = 3 (3 - x) - x (3 - x)$

Apply the distributive property.

$x - 5 = 3 \cdot 3 + 3 (- x) - x (3 - x)$

Apply the distributive property.

$x - 5 = 3 \cdot 3 + 3 (- x) - x \cdot 3 - x (- x)$

Simplify and combine like terms.

Simplify each term.

Multiply $3$ by $3$ .

$x - 5 = 9 + 3 (- x) - x \cdot 3 - x (- x)$

Multiply $- 1$ by $3$ .

$x - 5 = 9 - 3 x - x \cdot 3 - x (- x)$

Multiply $3$ by $- 1$ .

$x - 5 = 9 - 3 x - 3 x - x (- x)$

Rewrite using the commutative property of multiplication.

$x - 5 = 9 - 3 x - 3 x - 1 \cdot - 1 x \cdot x$

Multiply $x$ by $x$ by adding the exponents.

Move $x$ .

$x - 5 = 9 - 3 x - 3 x - 1 \cdot - 1 (x \cdot x)$

Multiply $x$ by $x$ .

$x - 5 = 9 - 3 x - 3 x - 1 \cdot - 1 x^{2}$

Multiply $- 1$ by $- 1$ .

$x - 5 = 9 - 3 x - 3 x + 1 x^{2}$

Multiply $x^{2}$ by $1$ .

$x - 5 = 9 - 3 x - 3 x + x^{2}$

Subtract $3 x$ from $- 3 x$ .

$x - 5 = 9 - 6 x + x^{2}$

Solve for $x$ .

Since $x$ is on the right side of the equation, switch the sides so it is on the left side of the equation.

$9 - 6 x + x^{2} = x - 5$

Move all terms containing $x$ to the left side of the equation.

Subtract $x$ from both sides of the equation.

$9 - 6 x + x^{2} - x = - 5$

Subtract $x$ from $- 6 x$ .

$9 + x^{2} - 7 x = - 5$

Move all terms to the left side of the equation and simplify.

Add $5$ to both sides of the equation.

$9 + x^{2} - 7 x + 5 = 0$

Add $9$ and $5$ .

$x^{2} - 7 x + 14 = 0$

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 1$ , $b = - 7$ , and $c = 14$ into the quadratic formula and solve for $x$ .

$\frac{7 \pm \sqrt{{(- 7)}^{2} - 4 \cdot (1 \cdot 14)}}{2 \cdot 1}$

Simplify.

Simplify the numerator.

Raise $- 7$ to the power of $2$ .

$x = \frac{7 \pm \sqrt{49 - 4 \cdot 1 \cdot 14}}{2 \cdot 1}$

Multiply $- 4$ by $1$ .

$x = \frac{7 \pm \sqrt{49 - 4 \cdot 14}}{2 \cdot 1}$

Multiply $- 4$ by $14$ .

$x = \frac{7 \pm \sqrt{49 - 56}}{2 \cdot 1}$

Subtract $56$ from $49$ .

$x = \frac{7 \pm \sqrt{- 7}}{2 \cdot 1}$

Rewrite $- 7$ as $- 1 (7)$ .

$x = \frac{7 \pm \sqrt{- 1 \cdot 7}}{2 \cdot 1}$

Rewrite $\sqrt{- 1 (7)}$ as $\sqrt{- 1} \cdot \sqrt{7}$ .

$x = \frac{7 \pm \sqrt{- 1} \cdot \sqrt{7}}{2 \cdot 1}$

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \frac{7 \pm i \sqrt{7}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{7 \pm i \sqrt{7}}{2}$

The final answer is the combination of both solutions.

$x = \frac{7 + i \sqrt{7}}{2}, \frac{7 - i \sqrt{7}}{2}$

Solve for x square root of x-5+x=3

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