Solve by Factoring square root of 3x-5+ square root of 2x+3+1=0

$\sqrt{3 x - 5} + \sqrt{2 x + 3} + 1 = 0$

Step 1

Move all terms not containing $\sqrt{3 x - 5}$ to the right side of the equation.

Subtract $\sqrt{2 x + 3}$ from both sides of the equation.

$\sqrt{3 x - 5} + 1 = - \sqrt{2 x + 3}$

Subtract $1$ from both sides of the equation.

$\sqrt{3 x - 5} = - \sqrt{2 x + 3} - 1$

Step 2

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

${\sqrt{3 x - 5}}^{2} = {(- \sqrt{2 x + 3} - 1)}^{2}$

Step 3

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{3 x - 5}$ as ${(3 x - 5)}^{\frac{1}{2}}$ .

${({(3 x - 5)}^{\frac{1}{2}})}^{2} = {(- \sqrt{2 x + 3} - 1)}^{2}$

Simplify the left side.

Simplify ${({(3 x - 5)}^{\frac{1}{2}})}^{2}$ .

Multiply the exponents in ${({(3 x - 5)}^{\frac{1}{2}})}^{2}$ .

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

${(3 x - 5)}^{\frac{1}{2} \cdot 2} = {(- \sqrt{2 x + 3} - 1)}^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(3 x - 5)}^{\frac{1}{2} \cdot 2} = {(- \sqrt{2 x + 3} - 1)}^{2}$

Rewrite the expression.

${(3 x - 5)}^{1} = {(- \sqrt{2 x + 3} - 1)}^{2}$

Simplify.

$3 x - 5 = {(- \sqrt{2 x + 3} - 1)}^{2}$

Simplify the right side.

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

Rewrite ${(- \sqrt{2 x + 3} - 1)}^{2}$ as $(- \sqrt{2 x + 3} - 1) (- \sqrt{2 x + 3} - 1)$ .

$3 x - 5 = (- \sqrt{2 x + 3} - 1) (- \sqrt{2 x + 3} - 1)$

Expand $(- \sqrt{2 x + 3} - 1) (- \sqrt{2 x + 3} - 1)$ using the FOIL Method.

Apply the distributive property.

$3 x - 5 = - \sqrt{2 x + 3} (- \sqrt{2 x + 3} - 1) - 1 (- \sqrt{2 x + 3} - 1)$

Apply the distributive property.

$3 x - 5 = - \sqrt{2 x + 3} (- \sqrt{2 x + 3}) - \sqrt{2 x + 3} \cdot - 1 - 1 (- \sqrt{2 x + 3} - 1)$

Apply the distributive property.

$3 x - 5 = - \sqrt{2 x + 3} (- \sqrt{2 x + 3}) - \sqrt{2 x + 3} \cdot - 1 - 1 (- \sqrt{2 x + 3}) - 1 \cdot - 1$

Simplify and combine like terms.

Simplify each term.

Multiply $- \sqrt{2 x + 3} (- \sqrt{2 x + 3})$ .

Multiply $- 1$ by $- 1$ .

$3 x - 5 = 1 \sqrt{2 x + 3} \sqrt{2 x + 3} - \sqrt{2 x + 3} \cdot - 1 - 1 (- \sqrt{2 x + 3}) - 1 \cdot - 1$

Multiply $\sqrt{2 x + 3}$ by $1$ .

$3 x - 5 = \sqrt{2 x + 3} \sqrt{2 x + 3} - \sqrt{2 x + 3} \cdot - 1 - 1 (- \sqrt{2 x + 3}) - 1 \cdot - 1$

Raise $\sqrt{2 x + 3}$ to the power of $1$ .

$3 x - 5 = {\sqrt{2 x + 3}}^{1} \sqrt{2 x + 3} - \sqrt{2 x + 3} \cdot - 1 - 1 (- \sqrt{2 x + 3}) - 1 \cdot - 1$

Raise $\sqrt{2 x + 3}$ to the power of $1$ .

$3 x - 5 = {\sqrt{2 x + 3}}^{1} {\sqrt{2 x + 3}}^{1} - \sqrt{2 x + 3} \cdot - 1 - 1 (- \sqrt{2 x + 3}) - 1 \cdot - 1$

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

$3 x - 5 = {\sqrt{2 x + 3}}^{1 + 1} - \sqrt{2 x + 3} \cdot - 1 - 1 (- \sqrt{2 x + 3}) - 1 \cdot - 1$

Add $1$ and $1$ .

$3 x - 5 = {\sqrt{2 x + 3}}^{2} - \sqrt{2 x + 3} \cdot - 1 - 1 (- \sqrt{2 x + 3}) - 1 \cdot - 1$

Rewrite ${\sqrt{2 x + 3}}^{2}$ as $2 x + 3$ .

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

$3 x - 5 = {({(2 x + 3)}^{\frac{1}{2}})}^{2} - \sqrt{2 x + 3} \cdot - 1 - 1 (- \sqrt{2 x + 3}) - 1 \cdot - 1$

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

$3 x - 5 = {(2 x + 3)}^{\frac{1}{2} \cdot 2} - \sqrt{2 x + 3} \cdot - 1 - 1 (- \sqrt{2 x + 3}) - 1 \cdot - 1$

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

$3 x - 5 = {(2 x + 3)}^{\frac{2}{2}} - \sqrt{2 x + 3} \cdot - 1 - 1 (- \sqrt{2 x + 3}) - 1 \cdot - 1$

Cancel the common factor of $2$ .

Cancel the common factor.

$3 x - 5 = {(2 x + 3)}^{\frac{2}{2}} - \sqrt{2 x + 3} \cdot - 1 - 1 (- \sqrt{2 x + 3}) - 1 \cdot - 1$

Rewrite the expression.

$3 x - 5 = {(2 x + 3)}^{1} - \sqrt{2 x + 3} \cdot - 1 - 1 (- \sqrt{2 x + 3}) - 1 \cdot - 1$

Simplify.

$3 x - 5 = 2 x + 3 - \sqrt{2 x + 3} \cdot - 1 - 1 (- \sqrt{2 x + 3}) - 1 \cdot - 1$

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

Multiply $- 1$ by $- 1$ .

$3 x - 5 = 2 x + 3 + 1 \sqrt{2 x + 3} - 1 (- \sqrt{2 x + 3}) - 1 \cdot - 1$

Multiply $\sqrt{2 x + 3}$ by $1$ .

$3 x - 5 = 2 x + 3 + \sqrt{2 x + 3} - 1 (- \sqrt{2 x + 3}) - 1 \cdot - 1$

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

Multiply $- 1$ by $- 1$ .

$3 x - 5 = 2 x + 3 + \sqrt{2 x + 3} + 1 \sqrt{2 x + 3} - 1 \cdot - 1$

Multiply $\sqrt{2 x + 3}$ by $1$ .

$3 x - 5 = 2 x + 3 + \sqrt{2 x + 3} + \sqrt{2 x + 3} - 1 \cdot - 1$

Multiply $- 1$ by $- 1$ .

$3 x - 5 = 2 x + 3 + \sqrt{2 x + 3} + \sqrt{2 x + 3} + 1$

Add $3$ and $1$ .

$3 x - 5 = 2 x + 4 + \sqrt{2 x + 3} + \sqrt{2 x + 3}$

Add $\sqrt{2 x + 3}$ and $\sqrt{2 x + 3}$ .

$3 x - 5 = 2 x + 4 + 2 \sqrt{2 x + 3}$

Step 4

Solve for $2 \sqrt{2 x + 3}$ .

Rewrite the equation as $2 x + 4 + 2 \sqrt{2 x + 3} = 3 x - 5$ .

$2 x + 4 + 2 \sqrt{2 x + 3} = 3 x - 5$

Move all terms not containing $2 \sqrt{2 x + 3}$ to the right side of the equation.

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

$4 + 2 \sqrt{2 x + 3} = 3 x - 5 - 2 x$

Subtract $4$ from both sides of the equation.

$2 \sqrt{2 x + 3} = 3 x - 5 - 2 x - 4$

Subtract $2 x$ from $3 x$ .

$2 \sqrt{2 x + 3} = x - 5 - 4$

Subtract $4$ from $- 5$ .

$2 \sqrt{2 x + 3} = x - 9$

Step 5

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

${(2 \sqrt{2 x + 3})}^{2} = {(x - 9)}^{2}$

Step 6

Simplify each side of the equation.

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

${(2 {(2 x + 3)}^{\frac{1}{2}})}^{2} = {(x - 9)}^{2}$

Simplify the left side.

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

Apply the product rule to $2 {(2 x + 3)}^{\frac{1}{2}}$ .

$2^{2} {({(2 x + 3)}^{\frac{1}{2}})}^{2} = {(x - 9)}^{2}$

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

$4 {({(2 x + 3)}^{\frac{1}{2}})}^{2} = {(x - 9)}^{2}$

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

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

$4 {(2 x + 3)}^{\frac{1}{2} \cdot 2} = {(x - 9)}^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

$4 {(2 x + 3)}^{\frac{1}{2} \cdot 2} = {(x - 9)}^{2}$

Rewrite the expression.

$4 {(2 x + 3)}^{1} = {(x - 9)}^{2}$

Simplify.

$4 (2 x + 3) = {(x - 9)}^{2}$

Apply the distributive property.

$4 (2 x) + 4 \cdot 3 = {(x - 9)}^{2}$

Multiply.

Multiply $2$ by $4$ .

$8 x + 4 \cdot 3 = {(x - 9)}^{2}$

Multiply $4$ by $3$ .

$8 x + 12 = {(x - 9)}^{2}$

Simplify the right side.

Simplify ${(x - 9)}^{2}$ .

Rewrite ${(x - 9)}^{2}$ as $(x - 9) (x - 9)$ .

$8 x + 12 = (x - 9) (x - 9)$

Expand $(x - 9) (x - 9)$ using the FOIL Method.

Apply the distributive property.

$8 x + 12 = x (x - 9) - 9 (x - 9)$

Apply the distributive property.

$8 x + 12 = x \cdot x + x \cdot - 9 - 9 (x - 9)$

Apply the distributive property.

$8 x + 12 = x \cdot x + x \cdot - 9 - 9 x - 9 \cdot - 9$

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

$8 x + 12 = x^{2} + x \cdot - 9 - 9 x - 9 \cdot - 9$

Move $- 9$ to the left of $x$ .

$8 x + 12 = x^{2} - 9 \cdot x - 9 x - 9 \cdot - 9$

Multiply $- 9$ by $- 9$ .

$8 x + 12 = x^{2} - 9 x - 9 x + 81$

Subtract $9 x$ from $- 9 x$ .

$8 x + 12 = x^{2} - 18 x + 81$

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.

$x^{2} - 18 x + 81 = 8 x + 12$

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

Subtract $8 x$ from both sides of the equation.

$x^{2} - 18 x + 81 - 8 x = 12$

Subtract $8 x$ from $- 18 x$ .

$x^{2} - 26 x + 81 = 12$

Subtract $12$ from both sides of the equation.

$x^{2} - 26 x + 81 - 12 = 0$

Subtract $12$ from $81$ .

$x^{2} - 26 x + 69 = 0$

Factor $x^{2} - 26 x + 69$ 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 $69$ and whose sum is $- 26$ .

$- 23, - 3$

Write the factored form using these integers.

$(x - 23) (x - 3) = 0$

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

$x - 23 = 0$

$x - 3 = 0$

Set $x - 23$ equal to $0$ and solve for $x$ .

Set $x - 23$ equal to $0$ .

$x - 23 = 0$

Add $23$ to both sides of the equation.

$x = 23$

Set $x - 3$ equal to $0$ and solve for $x$ .

Set $x - 3$ equal to $0$ .

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

The final solution is all the values that make $(x - 23) (x - 3) = 0$ true.

$x = 23, 3$

Step 8

Exclude the solutions that do not make $\sqrt{3 x - 5} + \sqrt{2 x + 3} + 1 = 0$ true.

$No solution$

Solve by Factoring square root of 3x-5+ square root of 2x+3+1=0

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