Solve for x square root of 4x+5=x-4

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

Step 1

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

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

Step 2

Simplify each side of the equation.

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

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

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, ${(a^{m})}^{n} = a^{m n}$ .

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

Simplify the right side.

Simplify ${(x - 4)}^{2}$ .

Rewrite ${(x - 4)}^{2}$ as $(x - 4) (x - 4)$ .

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

Expand $(x - 4) (x - 4)$ using the FOIL Method.

Apply the distributive property.

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

Apply the distributive property.

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

Apply the distributive property.

$4 x + 5 = x \cdot x + x \cdot - 4 - 4 x - 4 \cdot - 4$

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

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

Move $- 4$ to the left of $x$ .

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

Multiply $- 4$ by $- 4$ .

$4 x + 5 = x^{2} - 4 x - 4 x + 16$

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

$4 x + 5 = x^{2} - 8 x + 16$

Step 3

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.

$x^{2} - 8 x + 16 = 4 x + 5$

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

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

$x^{2} - 8 x + 16 - 4 x = 5$

Subtract $4 x$ from $- 8 x$ .

$x^{2} - 12 x + 16 = 5$

Subtract $5$ from both sides of the equation.

$x^{2} - 12 x + 16 - 5 = 0$

Subtract $5$ from $16$ .

$x^{2} - 12 x + 11 = 0$

Factor $x^{2} - 12 x + 11$ 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 $11$ and whose sum is $- 12$ .

$- 11, - 1$

Write the factored form using these integers.

$(x - 11) (x - 1) = 0$

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

$x - 11 = 0$

$x - 1 = 0$

Set $x - 11$ equal to $0$ and solve for $x$ .

Set $x - 11$ equal to $0$ .

$x - 11 = 0$

Add $11$ to both sides of the equation.

$x = 11$

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 $(x - 11) (x - 1) = 0$ true.

$x = 11, 1$

Step 4

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

$x = 11$

Solve for x square root of 4x+5=x-4

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