Solve for x square root of 2x^2-7=3-x

$\sqrt{2 x^{2} - 7} = 3 - x$

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

${\sqrt{2 x^{2} - 7}}^{2} = {(3 - x)}^{2}$

Simplify each side of the equation.

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

${({(2 x^{2} - 7)}^{\frac{1}{2}})}^{2} = {(3 - x)}^{2}$

Simplify the left side.

Simplify ${({(2 x^{2} - 7)}^{\frac{1}{2}})}^{2}$ .

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

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

${(2 x^{2} - 7)}^{\frac{1}{2} \cdot 2} = {(3 - x)}^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(2 x^{2} - 7)}^{\frac{1}{2} \cdot 2} = {(3 - x)}^{2}$

Rewrite the expression.

${(2 x^{2} - 7)}^{1} = {(3 - x)}^{2}$

Simplify.

$2 x^{2} - 7 = {(3 - x)}^{2}$

Simplify the right side.

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

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

$2 x^{2} - 7 = (3 - x) (3 - x)$

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

Apply the distributive property.

$2 x^{2} - 7 = 3 (3 - x) - x (3 - x)$

Apply the distributive property.

$2 x^{2} - 7 = 3 \cdot 3 + 3 (- x) - x (3 - x)$

Apply the distributive property.

$2 x^{2} - 7 = 3 \cdot 3 + 3 (- x) - x \cdot 3 - x (- x)$

Simplify and combine like terms.

Simplify each term.

Multiply $3$ by $3$ .

$2 x^{2} - 7 = 9 + 3 (- x) - x \cdot 3 - x (- x)$

Multiply $- 1$ by $3$ .

$2 x^{2} - 7 = 9 - 3 x - x \cdot 3 - x (- x)$

Multiply $3$ by $- 1$ .

$2 x^{2} - 7 = 9 - 3 x - 3 x - x (- x)$

Rewrite using the commutative property of multiplication.

$2 x^{2} - 7 = 9 - 3 x - 3 x - 1 \cdot - 1 x \cdot x$

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

Move $x$ .

$2 x^{2} - 7 = 9 - 3 x - 3 x - 1 \cdot - 1 (x \cdot x)$

Multiply $x$ by $x$ .

$2 x^{2} - 7 = 9 - 3 x - 3 x - 1 \cdot - 1 x^{2}$

Multiply $- 1$ by $- 1$ .

$2 x^{2} - 7 = 9 - 3 x - 3 x + 1 x^{2}$

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

$2 x^{2} - 7 = 9 - 3 x - 3 x + x^{2}$

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

$2 x^{2} - 7 = 9 - 6 x + x^{2}$

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.

$9 - 6 x + x^{2} = 2 x^{2} - 7$

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

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

$9 - 6 x + x^{2} - 2 x^{2} = - 7$

Subtract $2 x^{2}$ from $x^{2}$ .

$9 - 6 x - x^{2} = - 7$

Add $7$ to both sides of the equation.

$9 - 6 x - x^{2} + 7 = 0$

Add $9$ and $7$ .

$- 6 x - x^{2} + 16 = 0$

Factor the left side of the equation.

Factor $- 1$ out of $- 6 x - x^{2} + 16$ .

Reorder $- 6 x$ and $- x^{2}$ .

$- x^{2} - 6 x + 16 = 0$

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

$- (x^{2}) - 6 x + 16 = 0$

Factor $- 1$ out of $- 6 x$ .

$- (x^{2}) - (6 x) + 16 = 0$

Rewrite $16$ as $- 1 (- 16)$ .

$- (x^{2}) - (6 x) - 1 \cdot - 16 = 0$

Factor $- 1$ out of $- (x^{2}) - (6 x)$ .

$- (x^{2} + 6 x) - 1 \cdot - 16 = 0$

Factor $- 1$ out of $- (x^{2} + 6 x) - 1 (- 16)$ .

$- (x^{2} + 6 x - 16) = 0$

Factor.

Factor $x^{2} + 6 x - 16$ 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 $- 16$ and whose sum is $6$ .

$- 2, 8$

Write the factored form using these integers.

$- ((x - 2) (x + 8)) = 0$

Remove unnecessary parentheses.

$- (x - 2) (x + 8) = 0$

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

$x - 2 = 0$

$x + 8 = 0$

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

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

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

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 $- (x - 2) (x + 8) = 0$ true.

$x = 2, - 8$

Solve for x square root of 2x^2-7=3-x

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