Solve for x 2- square root of 2x+3=2x-1

$2 - \sqrt{2 x + 3} = 2 x - 1$

Step 1

Move all terms not containing $- \sqrt{2 x + 3}$ to the right side of the equation.

Subtract $2$ from both sides of the equation.

$- \sqrt{2 x + 3} = 2 x - 1 - 2$

Subtract $2$ from $- 1$ .

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

Step 2

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

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

Step 3

Simplify each side of the equation.

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

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

Simplify the left side.

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

Apply the product rule to $- {(2 x + 3)}^{\frac{1}{2}}$ .

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

Raise $- 1$ to the power of $2$ .

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

Multiply ${({(2 x + 3)}^{\frac{1}{2}})}^{2}$ by $1$ .

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

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

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

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

Simplify the right side.

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

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

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

Expand $(2 x - 3) (2 x - 3)$ using the FOIL Method.

Apply the distributive property.

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

Apply the distributive property.

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

Apply the distributive property.

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

Simplify and combine like terms.

Simplify each term.

Rewrite using the commutative property of multiplication.

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

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

Move $x$ .

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

Multiply $x$ by $x$ .

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

Multiply $2$ by $2$ .

$2 x + 3 = 4 x^{2} + 2 x \cdot - 3 - 3 (2 x) - 3 \cdot - 3$

Multiply $- 3$ by $2$ .

$2 x + 3 = 4 x^{2} - 6 x - 3 (2 x) - 3 \cdot - 3$

Multiply $2$ by $- 3$ .

$2 x + 3 = 4 x^{2} - 6 x - 6 x - 3 \cdot - 3$

Multiply $- 3$ by $- 3$ .

$2 x + 3 = 4 x^{2} - 6 x - 6 x + 9$

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

$2 x + 3 = 4 x^{2} - 12 x + 9$

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.

$4 x^{2} - 12 x + 9 = 2 x + 3$

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

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

$4 x^{2} - 12 x + 9 - 2 x = 3$

Subtract $2 x$ from $- 12 x$ .

$4 x^{2} - 14 x + 9 = 3$

Subtract $3$ from both sides of the equation.

$4 x^{2} - 14 x + 9 - 3 = 0$

Subtract $3$ from $9$ .

$4 x^{2} - 14 x + 6 = 0$

Factor the left side of the equation.

Factor $2$ out of $4 x^{2} - 14 x + 6$ .

Factor $2$ out of $4 x^{2}$ .

$2 (2 x^{2}) - 14 x + 6 = 0$

Factor $2$ out of $- 14 x$ .

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

Factor $2$ out of $6$ .

$2 (2 x^{2}) + 2 (- 7 x) + 2 (3) = 0$

Factor $2$ out of $2 (2 x^{2}) + 2 (- 7 x)$ .

$2 (2 x^{2} - 7 x) + 2 (3) = 0$

Factor $2$ out of $2 (2 x^{2} - 7 x) + 2 (3)$ .

$2 (2 x^{2} - 7 x + 3) = 0$

Factor.

Factor by grouping.

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 2 \cdot 3 = 6$ and whose sum is $b = - 7$ .

Factor $- 7$ out of $- 7 x$ .

$2 (2 x^{2} - 7 x + 3) = 0$

Rewrite $- 7$ as $- 1$ plus $- 6$

$2 (2 x^{2} + (- 1 - 6) x + 3) = 0$

Apply the distributive property.

$2 (2 x^{2} - 1 x - 6 x + 3) = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$2 ((2 x^{2} - 1 x) - 6 x + 3) = 0$

Factor out the greatest common factor (GCF) from each group.

$2 (x (2 x - 1) - 3 (2 x - 1)) = 0$

Factor the polynomial by factoring out the greatest common factor, $2 x - 1$ .

$2 ((2 x - 1) (x - 3)) = 0$

Remove unnecessary parentheses.

$2 (2 x - 1) (x - 3) = 0$

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

$2 x - 1 = 0$

$x - 3 = 0$

Set $2 x - 1$ equal to $0$ and solve for $x$ .

Set $2 x - 1$ equal to $0$ .

$2 x - 1 = 0$

Solve $2 x - 1 = 0$ for $x$ .

Add $1$ to both sides of the equation.

$2 x = 1$

Divide each term in $2 x = 1$ by $2$ and simplify.

Divide each term in $2 x = 1$ by $2$ .

$\frac{2 x}{2} = \frac{1}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{1}{2}$

Divide $x$ by $1$ .

$x = \frac{1}{2}$

Set $x - 3$ equal to $0$ and solve for $x$ .

Set $x - 3$ equal to $0$ .

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

The final solution is all the values that make $2 (2 x - 1) (x - 3) = 0$ true.

$x = \frac{1}{2}, 3$

Step 5

Exclude the solutions that do not make $2 - \sqrt{2 x + 3} = 2 x - 1$ true.

$x = \frac{1}{2}$

Step 6

The result can be shown in multiple forms.

Exact Form:

$x = \frac{1}{2}$

Decimal Form:

$x = 0.5$

Solve for x 2- square root of 2x+3=2x-1

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