Solve by Factoring square root of 4x+5-6=2x-11

$\sqrt{4 x + 5} - 6 = 2 x - 11$

Step 1

Move all the expressions to the left side of the equation.

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

$\sqrt{4 x + 5} - 6 - 2 x = - 11$

Add $11$ to both sides of the equation.

$\sqrt{4 x + 5} - 6 - 2 x + 11 = 0$

Step 2

Add $- 6$ and $11$ .

$\sqrt{4 x + 5} - 2 x + 5 = 0$

Step 3

Move all terms not containing $\sqrt{4 x + 5}$ to the right side of the equation.

Add $2 x$ to both sides of the equation.

$\sqrt{4 x + 5} + 5 = 2 x$

Subtract $5$ from both sides of the equation.

$\sqrt{4 x + 5} = 2 x - 5$

Step 4

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

${\sqrt{4 x + 5}}^{2} = {(2 x - 5)}^{2}$

Step 5

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{4 x + 5}$ as ${(4 x + 5)}^{\frac{1}{2}}$ .

${({(4 x + 5)}^{\frac{1}{2}})}^{2} = {(2 x - 5)}^{2}$

Simplify the left side.

Simplify ${({(4 x + 5)}^{\frac{1}{2}})}^{2}$ .

Multiply the exponents in ${({(4 x + 5)}^{\frac{1}{2}})}^{2}$ .

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

${(4 x + 5)}^{\frac{1}{2} \cdot 2} = {(2 x - 5)}^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(4 x + 5)}^{\frac{1}{2} \cdot 2} = {(2 x - 5)}^{2}$

Rewrite the expression.

${(4 x + 5)}^{1} = {(2 x - 5)}^{2}$

Simplify.

$4 x + 5 = {(2 x - 5)}^{2}$

Simplify the right side.

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

Rewrite ${(2 x - 5)}^{2}$ as $(2 x - 5) (2 x - 5)$ .

$4 x + 5 = (2 x - 5) (2 x - 5)$

Expand $(2 x - 5) (2 x - 5)$ using the FOIL Method.

Apply the distributive property.

$4 x + 5 = 2 x (2 x - 5) - 5 (2 x - 5)$

Apply the distributive property.

$4 x + 5 = 2 x (2 x) + 2 x \cdot - 5 - 5 (2 x - 5)$

Apply the distributive property.

$4 x + 5 = 2 x (2 x) + 2 x \cdot - 5 - 5 (2 x) - 5 \cdot - 5$

Simplify and combine like terms.

Simplify each term.

Rewrite using the commutative property of multiplication.

$4 x + 5 = 2 \cdot 2 x \cdot x + 2 x \cdot - 5 - 5 (2 x) - 5 \cdot - 5$

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

Move $x$ .

$4 x + 5 = 2 \cdot 2 (x \cdot x) + 2 x \cdot - 5 - 5 (2 x) - 5 \cdot - 5$

Multiply $x$ by $x$ .

$4 x + 5 = 2 \cdot 2 x^{2} + 2 x \cdot - 5 - 5 (2 x) - 5 \cdot - 5$

Multiply $2$ by $2$ .

$4 x + 5 = 4 x^{2} + 2 x \cdot - 5 - 5 (2 x) - 5 \cdot - 5$

Multiply $- 5$ by $2$ .

$4 x + 5 = 4 x^{2} - 10 x - 5 (2 x) - 5 \cdot - 5$

Multiply $2$ by $- 5$ .

$4 x + 5 = 4 x^{2} - 10 x - 10 x - 5 \cdot - 5$

Multiply $- 5$ by $- 5$ .

$4 x + 5 = 4 x^{2} - 10 x - 10 x + 25$

Subtract $10 x$ from $- 10 x$ .

$4 x + 5 = 4 x^{2} - 20 x + 25$

Step 6

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.

$4 x^{2} - 20 x + 25 = 4 x + 5$

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

Subtract $4 x$ from both sides of the equation.

$4 x^{2} - 20 x + 25 - 4 x = 5$

Subtract $4 x$ from $- 20 x$ .

$4 x^{2} - 24 x + 25 = 5$

Subtract $5$ from both sides of the equation.

$4 x^{2} - 24 x + 25 - 5 = 0$

Subtract $5$ from $25$ .

$4 x^{2} - 24 x + 20 = 0$

Factor the left side of the equation.

Factor $4$ out of $4 x^{2} - 24 x + 20$ .

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

$4 (x^{2}) - 24 x + 20 = 0$

Factor $4$ out of $- 24 x$ .

$4 (x^{2}) + 4 (- 6 x) + 20 = 0$

Factor $4$ out of $20$ .

$4 x^{2} + 4 (- 6 x) + 4 \cdot 5 = 0$

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

$4 (x^{2} - 6 x) + 4 \cdot 5 = 0$

Factor $4$ out of $4 (x^{2} - 6 x) + 4 \cdot 5$ .

$4 (x^{2} - 6 x + 5) = 0$

Factor.

Factor $x^{2} - 6 x + 5$ 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 $5$ and whose sum is $- 6$ .

$- 5, - 1$

Write the factored form using these integers.

$4 ((x - 5) (x - 1)) = 0$

Remove unnecessary parentheses.

$4 (x - 5) (x - 1) = 0$

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

$x - 5 = 0$

$x - 1 = 0$

Set $x - 5$ equal to $0$ and solve for $x$ .

Set $x - 5$ equal to $0$ .

$x - 5 = 0$

Add $5$ to both sides of the equation.

$x = 5$

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

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

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

The final solution is all the values that make $4 (x - 5) (x - 1) = 0$ true.

$x = 5, 1$

Step 7

Exclude the solutions that do not make $\sqrt{4 x + 5} - 2 x + 5 = 0$ true.

$x = 5$

Solve by Factoring square root of 4x+5-6=2x-11

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