Solve for x square root of 4x+1 = square root of 5x+4

$\sqrt{4 x} + 1 = \sqrt{5 x + 4}$

Subtract $1$ from both sides of the equation.

$\sqrt{4 x} = \sqrt{5 x + 4} - 1$

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

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

Simplify each side of the equation.

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

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

Simplify the left side.

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

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

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

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

Simplify the right side.

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

Rewrite ${(\sqrt{5 x + 4} - 1)}^{2}$ as $(\sqrt{5 x + 4} - 1) (\sqrt{5 x + 4} - 1)$ .

$4 x = (\sqrt{5 x + 4} - 1) (\sqrt{5 x + 4} - 1)$

Expand $(\sqrt{5 x + 4} - 1) (\sqrt{5 x + 4} - 1)$ using the FOIL Method.

Apply the distributive property.

$4 x = \sqrt{5 x + 4} (\sqrt{5 x + 4} - 1) - 1 (\sqrt{5 x + 4} - 1)$

Apply the distributive property.

$4 x = \sqrt{5 x + 4} \sqrt{5 x + 4} + \sqrt{5 x + 4} \cdot - 1 - 1 (\sqrt{5 x + 4} - 1)$

Apply the distributive property.

$4 x = \sqrt{5 x + 4} \sqrt{5 x + 4} + \sqrt{5 x + 4} \cdot - 1 - 1 \sqrt{5 x + 4} - 1 \cdot - 1$

Simplify and combine like terms.

Simplify each term.

Multiply $\sqrt{5 x + 4} \sqrt{5 x + 4}$ .

Raise $\sqrt{5 x + 4}$ to the power of $1$ .

$4 x = {\sqrt{5 x + 4}}^{1} \sqrt{5 x + 4} + \sqrt{5 x + 4} \cdot - 1 - 1 \sqrt{5 x + 4} - 1 \cdot - 1$

Raise $\sqrt{5 x + 4}$ to the power of $1$ .

$4 x = {\sqrt{5 x + 4}}^{1} {\sqrt{5 x + 4}}^{1} + \sqrt{5 x + 4} \cdot - 1 - 1 \sqrt{5 x + 4} - 1 \cdot - 1$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$4 x = {\sqrt{5 x + 4}}^{1 + 1} + \sqrt{5 x + 4} \cdot - 1 - 1 \sqrt{5 x + 4} - 1 \cdot - 1$

Add $1$ and $1$ .

$4 x = {\sqrt{5 x + 4}}^{2} + \sqrt{5 x + 4} \cdot - 1 - 1 \sqrt{5 x + 4} - 1 \cdot - 1$

Rewrite ${\sqrt{5 x + 4}}^{2}$ as $5 x + 4$ .

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

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

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

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

Combine $\frac{1}{2}$ and $2$ .

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

$4 x = 5 x + 4 + \sqrt{5 x + 4} \cdot - 1 - 1 \sqrt{5 x + 4} - 1 \cdot - 1$

Move $- 1$ to the left of $\sqrt{5 x + 4}$ .

$4 x = 5 x + 4 - 1 \cdot \sqrt{5 x + 4} - 1 \sqrt{5 x + 4} - 1 \cdot - 1$

Rewrite $- 1 \sqrt{5 x + 4}$ as $- \sqrt{5 x + 4}$ .

$4 x = 5 x + 4 - \sqrt{5 x + 4} - 1 \sqrt{5 x + 4} - 1 \cdot - 1$

Rewrite $- 1 \sqrt{5 x + 4}$ as $- \sqrt{5 x + 4}$ .

$4 x = 5 x + 4 - \sqrt{5 x + 4} - \sqrt{5 x + 4} - 1 \cdot - 1$

Multiply $- 1$ by $- 1$ .

$4 x = 5 x + 4 - \sqrt{5 x + 4} - \sqrt{5 x + 4} + 1$

Add $4$ and $1$ .

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

Subtract $\sqrt{5 x + 4}$ from $- \sqrt{5 x + 4}$ .

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

Solve for $- 2 \sqrt{5 x + 4}$ .

Rewrite the equation as $5 x + 5 - 2 \sqrt{5 x + 4} = 4 x$ .

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

Move all terms not containing $- 2 \sqrt{5 x + 4}$ to the right side of the equation.

Subtract $5 x$ from both sides of the equation.

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

Subtract $5$ from both sides of the equation.

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

Subtract $5 x$ from $4 x$ .

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

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

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

Simplify each side of the equation.

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

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

Simplify the left side.

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

Apply the product rule to $- 2 {(5 x + 4)}^{\frac{1}{2}}$ .

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

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

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

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

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

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

Apply the distributive property.

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

Multiply.

Multiply $5$ by $4$ .

$20 x + 4 \cdot 4 = {(- x - 5)}^{2}$

Multiply $4$ by $4$ .

$20 x + 16 = {(- x - 5)}^{2}$

Simplify the right side.

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

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

$20 x + 16 = (- x - 5) (- x - 5)$

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

Apply the distributive property.

$20 x + 16 = - x (- x - 5) - 5 (- x - 5)$

Apply the distributive property.

$20 x + 16 = - x (- x) - x \cdot - 5 - 5 (- x - 5)$

Apply the distributive property.

$20 x + 16 = - x (- x) - x \cdot - 5 - 5 (- x) - 5 \cdot - 5$

Simplify and combine like terms.

Simplify each term.

Rewrite using the commutative property of multiplication.

$20 x + 16 = - 1 \cdot - 1 x \cdot x - x \cdot - 5 - 5 (- x) - 5 \cdot - 5$

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

Move $x$ .

$20 x + 16 = - 1 \cdot - 1 (x \cdot x) - x \cdot - 5 - 5 (- x) - 5 \cdot - 5$

Multiply $x$ by $x$ .

$20 x + 16 = - 1 \cdot - 1 x^{2} - x \cdot - 5 - 5 (- x) - 5 \cdot - 5$

Multiply $- 1$ by $- 1$ .

$20 x + 16 = 1 x^{2} - x \cdot - 5 - 5 (- x) - 5 \cdot - 5$

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

$20 x + 16 = x^{2} - x \cdot - 5 - 5 (- x) - 5 \cdot - 5$

Multiply $- 5$ by $- 1$ .

$20 x + 16 = x^{2} + 5 x - 5 (- x) - 5 \cdot - 5$

Multiply $- 1$ by $- 5$ .

$20 x + 16 = x^{2} + 5 x + 5 x - 5 \cdot - 5$

Multiply $- 5$ by $- 5$ .

$20 x + 16 = x^{2} + 5 x + 5 x + 25$

Add $5 x$ and $5 x$ .

$20 x + 16 = x^{2} + 10 x + 25$

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.

$x^{2} + 10 x + 25 = 20 x + 16$

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

Subtract $20 x$ from both sides of the equation.

$x^{2} + 10 x + 25 - 20 x = 16$

Subtract $20 x$ from $10 x$ .

$x^{2} - 10 x + 25 = 16$

Subtract $16$ from both sides of the equation.

$x^{2} - 10 x + 25 - 16 = 0$

Subtract $16$ from $25$ .

$x^{2} - 10 x + 9 = 0$

Factor $x^{2} - 10 x + 9$ 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 $9$ and whose sum is $- 10$ .

$- 9, - 1$

Write the factored form using these integers.

$(x - 9) (x - 1) = 0$

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

$x - 9 = 0$

$x - 1 = 0$

Set $x - 9$ equal to $0$ and solve for $x$ .

Set $x - 9$ equal to $0$ .

$x - 9 = 0$

Add $9$ to both sides of the equation.

$x = 9$

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 $(x - 9) (x - 1) = 0$ true.

$x = 9, 1$

Solve for x square root of 4x+1 = square root of 5x+4

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