Solve by Factoring square root of 7x+1=1+ square root of 5x

$\sqrt{7 x + 1} = 1 + \sqrt{5 x}$

Step 1

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

Subtract $1$ from both sides of the equation.

$\sqrt{7 x + 1} - 1 = \sqrt{5 x}$

Subtract $\sqrt{5 x}$ from both sides of the equation.

$\sqrt{7 x + 1} - 1 - \sqrt{5 x} = 0$

Step 2

Move all terms not containing $\sqrt{7 x + 1}$ to the right side of the equation.

Add $1$ to both sides of the equation.

$\sqrt{7 x + 1} - \sqrt{5 x} = 1$

Add $\sqrt{5 x}$ to both sides of the equation.

$\sqrt{7 x + 1} = 1 + \sqrt{5 x}$

Step 3

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

${\sqrt{7 x + 1}}^{2} = {(1 + \sqrt{5 x})}^{2}$

Step 4

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{7 x + 1}$ as ${(7 x + 1)}^{\frac{1}{2}}$ .

${({(7 x + 1)}^{\frac{1}{2}})}^{2} = {(1 + \sqrt{5 x})}^{2}$

Simplify the left side.

Simplify ${({(7 x + 1)}^{\frac{1}{2}})}^{2}$ .

Multiply the exponents in ${({(7 x + 1)}^{\frac{1}{2}})}^{2}$ .

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

${(7 x + 1)}^{\frac{1}{2} \cdot 2} = {(1 + \sqrt{5 x})}^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(7 x + 1)}^{\frac{1}{2} \cdot 2} = {(1 + \sqrt{5 x})}^{2}$

Rewrite the expression.

${(7 x + 1)}^{1} = {(1 + \sqrt{5 x})}^{2}$

Simplify.

$7 x + 1 = {(1 + \sqrt{5 x})}^{2}$

Simplify the right side.

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

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

$7 x + 1 = (1 + \sqrt{5 x}) (1 + \sqrt{5 x})$

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

Apply the distributive property.

$7 x + 1 = 1 (1 + \sqrt{5 x}) + \sqrt{5 x} (1 + \sqrt{5 x})$

Apply the distributive property.

$7 x + 1 = 1 \cdot 1 + 1 \sqrt{5 x} + \sqrt{5 x} (1 + \sqrt{5 x})$

Apply the distributive property.

$7 x + 1 = 1 \cdot 1 + 1 \sqrt{5 x} + \sqrt{5 x} \cdot 1 + \sqrt{5 x} \sqrt{5 x}$

Simplify and combine like terms.

Simplify each term.

Multiply $1$ by $1$ .

$7 x + 1 = 1 + 1 \sqrt{5 x} + \sqrt{5 x} \cdot 1 + \sqrt{5 x} \sqrt{5 x}$

Multiply $\sqrt{5 x}$ by $1$ .

$7 x + 1 = 1 + \sqrt{5 x} + \sqrt{5 x} \cdot 1 + \sqrt{5 x} \sqrt{5 x}$

Multiply $\sqrt{5 x}$ by $1$ .

$7 x + 1 = 1 + \sqrt{5 x} + \sqrt{5 x} + \sqrt{5 x} \sqrt{5 x}$

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

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

$7 x + 1 = 1 + \sqrt{5 x} + \sqrt{5 x} + {\sqrt{5 x}}^{1} \sqrt{5 x}$

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

$7 x + 1 = 1 + \sqrt{5 x} + \sqrt{5 x} + {\sqrt{5 x}}^{1} {\sqrt{5 x}}^{1}$

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

$7 x + 1 = 1 + \sqrt{5 x} + \sqrt{5 x} + {\sqrt{5 x}}^{1 + 1}$

Add $1$ and $1$ .

$7 x + 1 = 1 + \sqrt{5 x} + \sqrt{5 x} + {\sqrt{5 x}}^{2}$

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

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

$7 x + 1 = 1 + \sqrt{5 x} + \sqrt{5 x} + {({(5 x)}^{\frac{1}{2}})}^{2}$

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

$7 x + 1 = 1 + \sqrt{5 x} + \sqrt{5 x} + {(5 x)}^{\frac{1}{2} \cdot 2}$

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

$7 x + 1 = 1 + \sqrt{5 x} + \sqrt{5 x} + {(5 x)}^{\frac{2}{2}}$

Cancel the common factor of $2$ .

Cancel the common factor.

$7 x + 1 = 1 + \sqrt{5 x} + \sqrt{5 x} + {(5 x)}^{\frac{2}{2}}$

Rewrite the expression.

$7 x + 1 = 1 + \sqrt{5 x} + \sqrt{5 x} + {(5 x)}^{1}$

Simplify.

$7 x + 1 = 1 + \sqrt{5 x} + \sqrt{5 x} + 5 x$

Add $\sqrt{5 x}$ and $\sqrt{5 x}$ .

$7 x + 1 = 1 + 2 \sqrt{5 x} + 5 x$

Step 5

Solve for $2 \sqrt{5 x}$ .

Rewrite the equation as $1 + 2 \sqrt{5 x} + 5 x = 7 x + 1$ .

$1 + 2 \sqrt{5 x} + 5 x = 7 x + 1$

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

Subtract $1$ from both sides of the equation.

$2 \sqrt{5 x} + 5 x = 7 x + 1 - 1$

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

$2 \sqrt{5 x} = 7 x + 1 - 1 - 5 x$

Combine the opposite terms in $7 x + 1 - 1 - 5 x$ .

Subtract $1$ from $1$ .

$2 \sqrt{5 x} = 7 x + 0 - 5 x$

Add $7 x$ and $0$ .

$2 \sqrt{5 x} = 7 x - 5 x$

Subtract $5 x$ from $7 x$ .

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

Step 6

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

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

Step 7

Simplify each side of the equation.

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

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

Simplify the left side.

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

Apply the product rule to $5 x$ .

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

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

Apply the product rule to $2 \cdot 5^{\frac{1}{2}} x^{\frac{1}{2}}$ .

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

Apply the product rule to $2 \cdot 5^{\frac{1}{2}}$ .

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

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

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

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

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

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Evaluate the exponent.

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

Multiply $4$ by $5$ .

$20 {(x^{\frac{1}{2}})}^{2} = {(2 x)}^{2}$

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

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

$20 x^{\frac{1}{2} \cdot 2} = {(2 x)}^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

$20 x^{\frac{1}{2} \cdot 2} = {(2 x)}^{2}$

Rewrite the expression.

$20 x^{1} = {(2 x)}^{2}$

Simplify.

$20 x = {(2 x)}^{2}$

Simplify the right side.

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

Apply the product rule to $2 x$ .

$20 x = 2^{2} x^{2}$

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

$20 x = 4 x^{2}$

Step 8

Solve for $x$ .

Subtract $4 x^{2}$ from both sides of the equation.

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

Factor the left side of the equation.

Let $u = x$ . Substitute $u$ for all occurrences of $x$ .

$20 u - 4 u^{2} = 0$

Factor $4 u$ out of $20 u - 4 u^{2}$ .

Factor $4 u$ out of $20 u$ .

$4 u (5) - 4 u^{2} = 0$

Factor $4 u$ out of $- 4 u^{2}$ .

$4 u (5) + 4 u (- u) = 0$

Factor $4 u$ out of $4 u (5) + 4 u (- u)$ .

$4 u (5 - u) = 0$

Replace all occurrences of $u$ with $x$ .

$4 x (5 - x) = 0$

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

$x = 0$

$5 - x = 0$

Set $x$ equal to $0$ .

$x = 0$

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

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

$5 - x = 0$

Solve $5 - x = 0$ for $x$ .

Subtract $5$ from both sides of the equation.

$- x = - 5$

Divide each term in $- x = - 5$ by $- 1$ and simplify.

Divide each term in $- x = - 5$ by $- 1$ .

$\frac{- x}{- 1} = \frac{- 5}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{- 5}{- 1}$

Divide $x$ by $1$ .

$x = \frac{- 5}{- 1}$

Simplify the right side.

Divide $- 5$ by $- 1$ .

$x = 5$

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

$x = 0, 5$

Solve by Factoring square root of 7x+1=1+ square root of 5x

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