Solve for x square root of -2x+4+ square root of 2x+5=3

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

Step 1

Subtract $\sqrt{2 x + 5}$ from both sides of the equation.

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

Step 2

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

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

Step 3

Simplify each side of the equation.

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

${({(- 2 x + 4)}^{\frac{1}{2}})}^{2} = {(3 - \sqrt{2 x + 5})}^{2}$

Simplify the left side.

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

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

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

${(- 2 x + 4)}^{\frac{1}{2} \cdot 2} = {(3 - \sqrt{2 x + 5})}^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(- 2 x + 4)}^{\frac{1}{2} \cdot 2} = {(3 - \sqrt{2 x + 5})}^{2}$

Rewrite the expression.

${(- 2 x + 4)}^{1} = {(3 - \sqrt{2 x + 5})}^{2}$

Simplify.

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

Simplify the right side.

Simplify ${(3 - \sqrt{2 x + 5})}^{2}$ .

Rewrite ${(3 - \sqrt{2 x + 5})}^{2}$ as $(3 - \sqrt{2 x + 5}) (3 - \sqrt{2 x + 5})$ .

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

Expand $(3 - \sqrt{2 x + 5}) (3 - \sqrt{2 x + 5})$ using the FOIL Method.

Apply the distributive property.

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

Apply the distributive property.

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

Apply the distributive property.

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

Simplify and combine like terms.

Simplify each term.

Multiply $3$ by $3$ .

$- 2 x + 4 = 9 + 3 (- \sqrt{2 x + 5}) - \sqrt{2 x + 5} \cdot 3 - \sqrt{2 x + 5} (- \sqrt{2 x + 5})$

Multiply $- 1$ by $3$ .

$- 2 x + 4 = 9 - 3 \sqrt{2 x + 5} - \sqrt{2 x + 5} \cdot 3 - \sqrt{2 x + 5} (- \sqrt{2 x + 5})$

Multiply $3$ by $- 1$ .

$- 2 x + 4 = 9 - 3 \sqrt{2 x + 5} - 3 \sqrt{2 x + 5} - \sqrt{2 x + 5} (- \sqrt{2 x + 5})$

Multiply $- \sqrt{2 x + 5} (- \sqrt{2 x + 5})$ .

Multiply $- 1$ by $- 1$ .

$- 2 x + 4 = 9 - 3 \sqrt{2 x + 5} - 3 \sqrt{2 x + 5} + 1 \sqrt{2 x + 5} \sqrt{2 x + 5}$

Multiply $\sqrt{2 x + 5}$ by $1$ .

$- 2 x + 4 = 9 - 3 \sqrt{2 x + 5} - 3 \sqrt{2 x + 5} + \sqrt{2 x + 5} \sqrt{2 x + 5}$

Raise $\sqrt{2 x + 5}$ to the power of $1$ .

$- 2 x + 4 = 9 - 3 \sqrt{2 x + 5} - 3 \sqrt{2 x + 5} + {\sqrt{2 x + 5}}^{1} \sqrt{2 x + 5}$

Raise $\sqrt{2 x + 5}$ to the power of $1$ .

$- 2 x + 4 = 9 - 3 \sqrt{2 x + 5} - 3 \sqrt{2 x + 5} + {\sqrt{2 x + 5}}^{1} {\sqrt{2 x + 5}}^{1}$

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

$- 2 x + 4 = 9 - 3 \sqrt{2 x + 5} - 3 \sqrt{2 x + 5} + {\sqrt{2 x + 5}}^{1 + 1}$

Add $1$ and $1$ .

$- 2 x + 4 = 9 - 3 \sqrt{2 x + 5} - 3 \sqrt{2 x + 5} + {\sqrt{2 x + 5}}^{2}$

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

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

$- 2 x + 4 = 9 - 3 \sqrt{2 x + 5} - 3 \sqrt{2 x + 5} + {({(2 x + 5)}^{\frac{1}{2}})}^{2}$

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

$- 2 x + 4 = 9 - 3 \sqrt{2 x + 5} - 3 \sqrt{2 x + 5} + {(2 x + 5)}^{\frac{1}{2} \cdot 2}$

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

$- 2 x + 4 = 9 - 3 \sqrt{2 x + 5} - 3 \sqrt{2 x + 5} + {(2 x + 5)}^{\frac{2}{2}}$

Cancel the common factor of $2$ .

Cancel the common factor.

$- 2 x + 4 = 9 - 3 \sqrt{2 x + 5} - 3 \sqrt{2 x + 5} + {(2 x + 5)}^{\frac{2}{2}}$

Rewrite the expression.

$- 2 x + 4 = 9 - 3 \sqrt{2 x + 5} - 3 \sqrt{2 x + 5} + {(2 x + 5)}^{1}$

Simplify.

$- 2 x + 4 = 9 - 3 \sqrt{2 x + 5} - 3 \sqrt{2 x + 5} + 2 x + 5$

Add $9$ and $5$ .

$- 2 x + 4 = 14 - 3 \sqrt{2 x + 5} - 3 \sqrt{2 x + 5} + 2 x$

Subtract $3 \sqrt{2 x + 5}$ from $- 3 \sqrt{2 x + 5}$ .

$- 2 x + 4 = 14 - 6 \sqrt{2 x + 5} + 2 x$

Step 4

Solve for $- 6 \sqrt{2 x + 5}$ .

Rewrite the equation as $14 - 6 \sqrt{2 x + 5} + 2 x = - 2 x + 4$ .

$14 - 6 \sqrt{2 x + 5} + 2 x = - 2 x + 4$

Move all terms not containing $- 6 \sqrt{2 x + 5}$ to the right side of the equation.

Subtract $14$ from both sides of the equation.

$- 6 \sqrt{2 x + 5} + 2 x = - 2 x + 4 - 14$

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

$- 6 \sqrt{2 x + 5} = - 2 x + 4 - 14 - 2 x$

Subtract $2 x$ from $- 2 x$ .

$- 6 \sqrt{2 x + 5} = - 4 x + 4 - 14$

Subtract $14$ from $4$ .

$- 6 \sqrt{2 x + 5} = - 4 x - 10$

Step 5

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

${(- 6 \sqrt{2 x + 5})}^{2} = {(- 4 x - 10)}^{2}$

Step 6

Simplify each side of the equation.

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

${(- 6 {(2 x + 5)}^{\frac{1}{2}})}^{2} = {(- 4 x - 10)}^{2}$

Simplify the left side.

Simplify ${(- 6 {(2 x + 5)}^{\frac{1}{2}})}^{2}$ .

Apply the product rule to $- 6 {(2 x + 5)}^{\frac{1}{2}}$ .

${(- 6)}^{2} {({(2 x + 5)}^{\frac{1}{2}})}^{2} = {(- 4 x - 10)}^{2}$

Raise $- 6$ to the power of $2$ .

$36 {({(2 x + 5)}^{\frac{1}{2}})}^{2} = {(- 4 x - 10)}^{2}$

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

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

$36 {(2 x + 5)}^{\frac{1}{2} \cdot 2} = {(- 4 x - 10)}^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

$36 {(2 x + 5)}^{\frac{1}{2} \cdot 2} = {(- 4 x - 10)}^{2}$

Rewrite the expression.

$36 {(2 x + 5)}^{1} = {(- 4 x - 10)}^{2}$

Simplify.

$36 (2 x + 5) = {(- 4 x - 10)}^{2}$

Apply the distributive property.

$36 (2 x) + 36 \cdot 5 = {(- 4 x - 10)}^{2}$

Multiply.

Multiply $2$ by $36$ .

$72 x + 36 \cdot 5 = {(- 4 x - 10)}^{2}$

Multiply $36$ by $5$ .

$72 x + 180 = {(- 4 x - 10)}^{2}$

Simplify the right side.

Simplify ${(- 4 x - 10)}^{2}$ .

Rewrite ${(- 4 x - 10)}^{2}$ as $(- 4 x - 10) (- 4 x - 10)$ .

$72 x + 180 = (- 4 x - 10) (- 4 x - 10)$

Expand $(- 4 x - 10) (- 4 x - 10)$ using the FOIL Method.

Apply the distributive property.

$72 x + 180 = - 4 x (- 4 x - 10) - 10 (- 4 x - 10)$

Apply the distributive property.

$72 x + 180 = - 4 x (- 4 x) - 4 x \cdot - 10 - 10 (- 4 x - 10)$

Apply the distributive property.

$72 x + 180 = - 4 x (- 4 x) - 4 x \cdot - 10 - 10 (- 4 x) - 10 \cdot - 10$

Simplify and combine like terms.

Simplify each term.

Rewrite using the commutative property of multiplication.

$72 x + 180 = - 4 \cdot - 4 x \cdot x - 4 x \cdot - 10 - 10 (- 4 x) - 10 \cdot - 10$

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

Move $x$ .

$72 x + 180 = - 4 \cdot - 4 (x \cdot x) - 4 x \cdot - 10 - 10 (- 4 x) - 10 \cdot - 10$

Multiply $x$ by $x$ .

$72 x + 180 = - 4 \cdot - 4 x^{2} - 4 x \cdot - 10 - 10 (- 4 x) - 10 \cdot - 10$

Multiply $- 4$ by $- 4$ .

$72 x + 180 = 16 x^{2} - 4 x \cdot - 10 - 10 (- 4 x) - 10 \cdot - 10$

Multiply $- 10$ by $- 4$ .

$72 x + 180 = 16 x^{2} + 40 x - 10 (- 4 x) - 10 \cdot - 10$

Multiply $- 4$ by $- 10$ .

$72 x + 180 = 16 x^{2} + 40 x + 40 x - 10 \cdot - 10$

Multiply $- 10$ by $- 10$ .

$72 x + 180 = 16 x^{2} + 40 x + 40 x + 100$

Add $40 x$ and $40 x$ .

$72 x + 180 = 16 x^{2} + 80 x + 100$

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.

$16 x^{2} + 80 x + 100 = 72 x + 180$

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

Subtract $72 x$ from both sides of the equation.

$16 x^{2} + 80 x + 100 - 72 x = 180$

Subtract $72 x$ from $80 x$ .

$16 x^{2} + 8 x + 100 = 180$

Subtract $180$ from both sides of the equation.

$16 x^{2} + 8 x + 100 - 180 = 0$

Subtract $180$ from $100$ .

$16 x^{2} + 8 x - 80 = 0$

Factor the left side of the equation.

Factor $8$ out of $16 x^{2} + 8 x - 80$ .

Factor $8$ out of $16 x^{2}$ .

$8 (2 x^{2}) + 8 x - 80 = 0$

Factor $8$ out of $8 x$ .

$8 (2 x^{2}) + 8 (x) - 80 = 0$

Factor $8$ out of $- 80$ .

$8 (2 x^{2}) + 8 x + 8 \cdot - 10 = 0$

Factor $8$ out of $8 (2 x^{2}) + 8 x$ .

$8 (2 x^{2} + x) + 8 \cdot - 10 = 0$

Factor $8$ out of $8 (2 x^{2} + x) + 8 \cdot - 10$ .

$8 (2 x^{2} + x - 10) = 0$

Factor.

Factor by grouping.

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 2 \cdot - 10 = - 20$ and whose sum is $b = 1$ .

Multiply by $1$ .

$8 (2 x^{2} + 1 x - 10) = 0$

Rewrite $1$ as $- 4$ plus $5$

$8 (2 x^{2} + (- 4 + 5) x - 10) = 0$

Apply the distributive property.

$8 (2 x^{2} - 4 x + 5 x - 10) = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$8 ((2 x^{2} - 4 x) + 5 x - 10) = 0$

Factor out the greatest common factor (GCF) from each group.

$8 (2 x (x - 2) + 5 (x - 2)) = 0$

Factor the polynomial by factoring out the greatest common factor, $x - 2$ .

$8 ((x - 2) (2 x + 5)) = 0$

Remove unnecessary parentheses.

$8 (x - 2) (2 x + 5) = 0$

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

$x - 2 = 0$

$2 x + 5 = 0$

Set $x - 2$ equal to $0$ and solve for $x$ .

Set $x - 2$ equal to $0$ .

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

Set $2 x + 5$ equal to $0$ and solve for $x$ .

Set $2 x + 5$ equal to $0$ .

$2 x + 5 = 0$

Solve $2 x + 5 = 0$ for $x$ .

Subtract $5$ from both sides of the equation.

$2 x = - 5$

Divide each term in $2 x = - 5$ by $2$ and simplify.

Divide each term in $2 x = - 5$ by $2$ .

$\frac{2 x}{2} = \frac{- 5}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{- 5}{2}$

Divide $x$ by $1$ .

$x = \frac{- 5}{2}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{5}{2}$

The final solution is all the values that make $8 (x - 2) (2 x + 5) = 0$ true.

$x = 2, - \frac{5}{2}$

Step 8

The result can be shown in multiple forms.

Exact Form:

$x = 2, - \frac{5}{2}$

Decimal Form:

$x = 2, - 2.5$

Mixed Number Form:

$x = 2, - 2 \frac{1}{2}$

Solve for x square root of -2x+4+ square root of 2x+5=3

Related Questions

Support LCMGCF.com by disabling your adblocker.