Solve by Factoring x+ square root of x=20

$x + \sqrt{x} = 20$

Step 1

Subtract $20$ from both sides of the equation.

$x + \sqrt{x} - 20 = 0$

Step 2

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

$x + x^{\frac{1}{2}} - 20 = 0$

Step 3

Rewrite $x$ as ${(x^{\frac{1}{2}})}^{2}$ .

${(x^{\frac{1}{2}})}^{2} + x^{\frac{1}{2}} - 20 = 0$

Step 4

Let $u = x^{\frac{1}{2}}$ . Substitute $u$ for all occurrences of $x^{\frac{1}{2}}$ .

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

Step 5

Factor $u^{2} + u - 20$ 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 $1$ .

$- 4, 5$

Write the factored form using these integers.

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

Step 6

Replace all occurrences of $u$ with $x^{\frac{1}{2}}$ .

$(x^{\frac{1}{2}} - 4) (x^{\frac{1}{2}} + 5) = 0$

Step 7

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

$x^{\frac{1}{2}} - 4 = 0$

$x^{\frac{1}{2}} + 5 = 0$

Step 8

Set $x^{\frac{1}{2}} - 4$ equal to $0$ and solve for $x$ .

Set $x^{\frac{1}{2}} - 4$ equal to $0$ .

$x^{\frac{1}{2}} - 4 = 0$

Solve $x^{\frac{1}{2}} - 4 = 0$ for $x$ .

Add $4$ to both sides of the equation.

$x^{\frac{1}{2}} = 4$

Raise each side of the equation to the power of $2$ to eliminate the fractional exponent on the left side.

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

Simplify the exponent.

Simplify the left side.

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

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

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

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

$x^{1} = 4^{2}$

Simplify.

$x = 4^{2}$

Simplify the right side.

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

$x = 16$

Step 9

Set $x^{\frac{1}{2}} + 5$ equal to $0$ and solve for $x$ .

Set $x^{\frac{1}{2}} + 5$ equal to $0$ .

$x^{\frac{1}{2}} + 5 = 0$

Solve $x^{\frac{1}{2}} + 5 = 0$ for $x$ .

Subtract $5$ from both sides of the equation.

$x^{\frac{1}{2}} = - 5$

Raise each side of the equation to the power of $2$ to eliminate the fractional exponent on the left side.

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

Simplify the exponent.

Simplify the left side.

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

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

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

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

Simplify the right side.

Raise $- 5$ to the power of $2$ .

$x = 25$

Step 10

The final solution is all the values that make $(x^{\frac{1}{2}} - 4) (x^{\frac{1}{2}} + 5) = 0$ true.

$x = 16, 25$

Step 11

Exclude the solutions that do not make $(x^{\frac{1}{2}} - 4) (x^{\frac{1}{2}} + 5) = 0$ true.

$x = 16$

Solve by Factoring x+ square root of x=20

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