Solve for x (x+5)^(1/2)=(5-2x)^(1/4)

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

Eliminate the fractional exponents by multiplying both exponents by the LCD.

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

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

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

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

Cancel the common factor of $2$ .

Factor $2$ out of $4$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

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

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

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

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

Cancel the common factor of $4$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

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

Add $2 x$ to both sides of the equation.

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

Simplify each term.

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

$(x + 5) (x + 5) + 2 x = 5$

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

Apply the distributive property.

$x (x + 5) + 5 (x + 5) + 2 x = 5$

Apply the distributive property.

$x \cdot x + x \cdot 5 + 5 (x + 5) + 2 x = 5$

Apply the distributive property.

$x \cdot x + x \cdot 5 + 5 x + 5 \cdot 5 + 2 x = 5$

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

$x^{2} + x \cdot 5 + 5 x + 5 \cdot 5 + 2 x = 5$

Move $5$ to the left of $x$ .

$x^{2} + 5 \cdot x + 5 x + 5 \cdot 5 + 2 x = 5$

Multiply $5$ by $5$ .

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

Add $5 x$ and $5 x$ .

$x^{2} + 10 x + 25 + 2 x = 5$

Add $10 x$ and $2 x$ .

$x^{2} + 12 x + 25 = 5$

Subtract $5$ from both sides of the equation.

$x^{2} + 12 x + 25 - 5 = 0$

Subtract $5$ from $25$ .

$x^{2} + 12 x + 20 = 0$

Factor $x^{2} + 12 x + 20$ 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 $20$ and whose sum is $12$ .

$2, 10$

Write the factored form using these integers.

$(x + 2) (x + 10) = 0$

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

$x + 2 = 0$

$x + 10 = 0$

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

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

$x + 2 = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

Set $x + 10$ equal to $0$ and solve for $x$ .

Set $x + 10$ equal to $0$ .

$x + 10 = 0$

Subtract $10$ from both sides of the equation.

$x = - 10$

The final solution is all the values that make $(x + 2) (x + 10) = 0$ true.

$x = - 2, - 10$

Exclude the solutions that do not make ${(x + 5)}^{\frac{1}{2}} = {(5 - 2 x)}^{\frac{1}{4}}$ true.

$x = - 2$

Solve for x (x+5)^(1/2)=(5-2x)^(1/4)

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