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

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

Step 1

Add $3$ to both sides of the equation.

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

Step 2

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

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

Step 3

Simplify each side of the equation.

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

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

Simplify the left side.

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

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

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

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

Simplify the right side.

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

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

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

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

Apply the distributive property.

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

Apply the distributive property.

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

Apply the distributive property.

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

Simplify and combine like terms.

Simplify each term.

Rewrite using the commutative property of multiplication.

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

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

Move $x$ .

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

Multiply $x$ by $x$ .

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

Multiply $2$ by $2$ .

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

Multiply $3$ by $2$ .

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

Multiply $2$ by $3$ .

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

Multiply $3$ by $3$ .

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

Add $6 x$ and $6 x$ .

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

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.

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

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

Subtract $x$ from both sides of the equation.

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

Subtract $x$ from $12 x$ .

$4 x^{2} + 11 x + 9 = 2$

Subtract $2$ from both sides of the equation.

$4 x^{2} + 11 x + 9 - 2 = 0$

Subtract $2$ from $9$ .

$4 x^{2} + 11 x + 7 = 0$

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 $a \cdot c = 4 \cdot 7 = 28$ and whose sum is $b = 11$ .

Factor $11$ out of $11 x$ .

$4 x^{2} + 11 (x) + 7 = 0$

Rewrite $11$ as $4$ plus $7$

$4 x^{2} + (4 + 7) x + 7 = 0$

Apply the distributive property.

$4 x^{2} + 4 x + 7 x + 7 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(4 x^{2} + 4 x) + 7 x + 7 = 0$

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

$4 x (x + 1) + 7 (x + 1) = 0$

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

$(x + 1) (4 x + 7) = 0$

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

$x + 1 = 0$

$4 x + 7 = 0$

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.

$x = - 1$

Set $4 x + 7$ equal to $0$ and solve for $x$ .

Set $4 x + 7$ equal to $0$ .

$4 x + 7 = 0$

Solve $4 x + 7 = 0$ for $x$ .

Subtract $7$ from both sides of the equation.

$4 x = - 7$

Divide each term in $4 x = - 7$ by $4$ and simplify.

Divide each term in $4 x = - 7$ by $4$ .

$\frac{4 x}{4} = \frac{- 7}{4}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 x}{4} = \frac{- 7}{4}$

Divide $x$ by $1$ .

$x = \frac{- 7}{4}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{7}{4}$

The final solution is all the values that make $(x + 1) (4 x + 7) = 0$ true.

$x = - 1, - \frac{7}{4}$

Step 5

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

$x = - 1$

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

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