Solve for x square root of 2x-3+x=3

$\sqrt{2 x - 3} + x = 3$

Step 1

Subtract $x$ from both sides of the equation.

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

Step 2

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

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

Step 3

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{2 x - 3}$ as ${(2 x - 3)}^{\frac{1}{2}}$ .

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

Simplify the left side.

Simplify ${({(2 x - 3)}^{\frac{1}{2}})}^{2}$ .

Multiply the exponents in ${({(2 x - 3)}^{\frac{1}{2}})}^{2}$ .

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

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

Simplify the right side.

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

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

$2 x - 3 = (3 - x) (3 - x)$

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

Apply the distributive property.

$2 x - 3 = 3 (3 - x) - x (3 - x)$

Apply the distributive property.

$2 x - 3 = 3 \cdot 3 + 3 (- x) - x (3 - x)$

Apply the distributive property.

$2 x - 3 = 3 \cdot 3 + 3 (- x) - x \cdot 3 - x (- x)$

Simplify and combine like terms.

Simplify each term.

Multiply $3$ by $3$ .

$2 x - 3 = 9 + 3 (- x) - x \cdot 3 - x (- x)$

Multiply $- 1$ by $3$ .

$2 x - 3 = 9 - 3 x - x \cdot 3 - x (- x)$

Multiply $3$ by $- 1$ .

$2 x - 3 = 9 - 3 x - 3 x - x (- x)$

Rewrite using the commutative property of multiplication.

$2 x - 3 = 9 - 3 x - 3 x - 1 \cdot - 1 x \cdot x$

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

Move $x$ .

$2 x - 3 = 9 - 3 x - 3 x - 1 \cdot - 1 (x \cdot x)$

Multiply $x$ by $x$ .

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

Multiply $- 1$ by $- 1$ .

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

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

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

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

$2 x - 3 = 9 - 6 x + x^{2}$

Step 4

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.

$9 - 6 x + x^{2} = 2 x - 3$

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

Subtract $2 x$ from both sides of the equation.

$9 - 6 x + x^{2} - 2 x = - 3$

Subtract $2 x$ from $- 6 x$ .

$9 + x^{2} - 8 x = - 3$

Add $3$ to both sides of the equation.

$9 + x^{2} - 8 x + 3 = 0$

Add $9$ and $3$ .

$x^{2} - 8 x + 12 = 0$

Factor $x^{2} - 8 x + 12$ 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 $12$ and whose sum is $- 8$ .

$- 6, - 2$

Write the factored form using these integers.

$(x - 6) (x - 2) = 0$

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

$x - 6 = 0$

$x - 2 = 0$

Set $x - 6$ equal to $0$ and solve for $x$ .

Set $x - 6$ equal to $0$ .

$x - 6 = 0$

Add $6$ to both sides of the equation.

$x = 6$

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.

$x = 2$

The final solution is all the values that make $(x - 6) (x - 2) = 0$ true.

$x = 6, 2$

Step 5

Exclude the solutions that do not make $\sqrt{2 x - 3} + x = 3$ true.

$x = 2$

Solve for x square root of 2x-3+x=3

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