Find All Complex Solutions square root of 3x+1- square root of x-1=2

$\sqrt{3 x + 1} - \sqrt{x - 1} = 2$

Step 1

Add $\sqrt{x - 1}$ to both sides of the equation.

$\sqrt{3 x + 1} = 2 + \sqrt{x - 1}$

Step 2

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

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

Step 3

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{3 x + 1}$ as ${(3 x + 1)}^{\frac{1}{2}}$ .

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

Simplify the left side.

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

Multiply the exponents in ${({(3 x + 1)}^{\frac{1}{2}})}^{2}$ .

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

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

Simplify the right side.

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

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

$3 x + 1 = (2 + \sqrt{x - 1}) (2 + \sqrt{x - 1})$

Expand $(2 + \sqrt{x - 1}) (2 + \sqrt{x - 1})$ using the FOIL Method.

Apply the distributive property.

$3 x + 1 = 2 (2 + \sqrt{x - 1}) + \sqrt{x - 1} (2 + \sqrt{x - 1})$

Apply the distributive property.

$3 x + 1 = 2 \cdot 2 + 2 \sqrt{x - 1} + \sqrt{x - 1} (2 + \sqrt{x - 1})$

Apply the distributive property.

$3 x + 1 = 2 \cdot 2 + 2 \sqrt{x - 1} + \sqrt{x - 1} \cdot 2 + \sqrt{x - 1} \sqrt{x - 1}$

Simplify and combine like terms.

Simplify each term.

Multiply $2$ by $2$ .

$3 x + 1 = 4 + 2 \sqrt{x - 1} + \sqrt{x - 1} \cdot 2 + \sqrt{x - 1} \sqrt{x - 1}$

Move $2$ to the left of $\sqrt{x - 1}$ .

$3 x + 1 = 4 + 2 \sqrt{x - 1} + 2 \cdot \sqrt{x - 1} + \sqrt{x - 1} \sqrt{x - 1}$

Multiply $\sqrt{x - 1} \sqrt{x - 1}$ .

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

$3 x + 1 = 4 + 2 \sqrt{x - 1} + 2 \sqrt{x - 1} + {\sqrt{x - 1}}^{1} \sqrt{x - 1}$

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

$3 x + 1 = 4 + 2 \sqrt{x - 1} + 2 \sqrt{x - 1} + {\sqrt{x - 1}}^{1} {\sqrt{x - 1}}^{1}$

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

$3 x + 1 = 4 + 2 \sqrt{x - 1} + 2 \sqrt{x - 1} + {\sqrt{x - 1}}^{1 + 1}$

Add $1$ and $1$ .

$3 x + 1 = 4 + 2 \sqrt{x - 1} + 2 \sqrt{x - 1} + {\sqrt{x - 1}}^{2}$

Rewrite ${\sqrt{x - 1}}^{2}$ as $x - 1$ .

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

$3 x + 1 = 4 + 2 \sqrt{x - 1} + 2 \sqrt{x - 1} + {({(x - 1)}^{\frac{1}{2}})}^{2}$

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

$3 x + 1 = 4 + 2 \sqrt{x - 1} + 2 \sqrt{x - 1} + {(x - 1)}^{\frac{1}{2} \cdot 2}$

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

$3 x + 1 = 4 + 2 \sqrt{x - 1} + 2 \sqrt{x - 1} + {(x - 1)}^{\frac{2}{2}}$

Cancel the common factor of $2$ .

Cancel the common factor.

$3 x + 1 = 4 + 2 \sqrt{x - 1} + 2 \sqrt{x - 1} + {(x - 1)}^{\frac{2}{2}}$

Rewrite the expression.

$3 x + 1 = 4 + 2 \sqrt{x - 1} + 2 \sqrt{x - 1} + {(x - 1)}^{1}$

Simplify.

$3 x + 1 = 4 + 2 \sqrt{x - 1} + 2 \sqrt{x - 1} + x - 1$

Subtract $1$ from $4$ .

$3 x + 1 = 3 + 2 \sqrt{x - 1} + 2 \sqrt{x - 1} + x$

Add $2 \sqrt{x - 1}$ and $2 \sqrt{x - 1}$ .

$3 x + 1 = 3 + 4 \sqrt{x - 1} + x$

Step 4

Solve for $4 \sqrt{x - 1}$ .

Rewrite the equation as $3 + 4 \sqrt{x - 1} + x = 3 x + 1$ .

$3 + 4 \sqrt{x - 1} + x = 3 x + 1$

Move all terms not containing $4 \sqrt{x - 1}$ to the right side of the equation.

Subtract $3$ from both sides of the equation.

$4 \sqrt{x - 1} + x = 3 x + 1 - 3$

Subtract $x$ from both sides of the equation.

$4 \sqrt{x - 1} = 3 x + 1 - 3 - x$

Subtract $x$ from $3 x$ .

$4 \sqrt{x - 1} = 2 x + 1 - 3$

Subtract $3$ from $1$ .

$4 \sqrt{x - 1} = 2 x - 2$

Step 5

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

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

Step 6

Simplify each side of the equation.

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

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

Simplify the left side.

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

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

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

Raise $4$ to the power of $2$ .

$16 {({(x - 1)}^{\frac{1}{2}})}^{2} = {(2 x - 2)}^{2}$

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

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

$16 {(x - 1)}^{\frac{1}{2} \cdot 2} = {(2 x - 2)}^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

$16 {(x - 1)}^{\frac{1}{2} \cdot 2} = {(2 x - 2)}^{2}$

Rewrite the expression.

$16 {(x - 1)}^{1} = {(2 x - 2)}^{2}$

Simplify.

$16 (x - 1) = {(2 x - 2)}^{2}$

Apply the distributive property.

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

Multiply $16$ by $- 1$ .

$16 x - 16 = {(2 x - 2)}^{2}$

Simplify the right side.

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

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

$16 x - 16 = (2 x - 2) (2 x - 2)$

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

Apply the distributive property.

$16 x - 16 = 2 x (2 x - 2) - 2 (2 x - 2)$

Apply the distributive property.

$16 x - 16 = 2 x (2 x) + 2 x \cdot - 2 - 2 (2 x - 2)$

Apply the distributive property.

$16 x - 16 = 2 x (2 x) + 2 x \cdot - 2 - 2 (2 x) - 2 \cdot - 2$

Simplify and combine like terms.

Simplify each term.

Rewrite using the commutative property of multiplication.

$16 x - 16 = 2 \cdot 2 x \cdot x + 2 x \cdot - 2 - 2 (2 x) - 2 \cdot - 2$

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

Move $x$ .

$16 x - 16 = 2 \cdot 2 (x \cdot x) + 2 x \cdot - 2 - 2 (2 x) - 2 \cdot - 2$

Multiply $x$ by $x$ .

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

Multiply $2$ by $2$ .

$16 x - 16 = 4 x^{2} + 2 x \cdot - 2 - 2 (2 x) - 2 \cdot - 2$

Multiply $- 2$ by $2$ .

$16 x - 16 = 4 x^{2} - 4 x - 2 (2 x) - 2 \cdot - 2$

Multiply $2$ by $- 2$ .

$16 x - 16 = 4 x^{2} - 4 x - 4 x - 2 \cdot - 2$

Multiply $- 2$ by $- 2$ .

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

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

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

Step 7

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.

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

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

Subtract $16 x$ from both sides of the equation.

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

Subtract $16 x$ from $- 8 x$ .

$4 x^{2} - 24 x + 4 = - 16$

Add $16$ to both sides of the equation.

$4 x^{2} - 24 x + 4 + 16 = 0$

Add $4$ and $16$ .

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

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.

$x = 5, 1$

Find All Complex Solutions square root of 3x+1- square root of x-1=2

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