Solve for x 2 square root of x+7=5

$2 \sqrt{x} + 7 = 5$

Step 1

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

Subtract $7$ from both sides of the equation.

$2 \sqrt{x} = 5 - 7$

Subtract $7$ from $5$ .

$2 \sqrt{x} = - 2$

Step 2

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

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

Step 3

Simplify each side of the equation.

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

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

Simplify the left side.

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

Apply the product rule to $2 x^{\frac{1}{2}}$ .

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

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

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

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

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

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

$4 x = {(- 2)}^{2}$

Simplify the right side.

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

$4 x = 4$

Step 4

Divide each term in $4 x = 4$ by $4$ and simplify.

Divide each term in $4 x = 4$ by $4$ .

$\frac{4 x}{4} = \frac{4}{4}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 x}{4} = \frac{4}{4}$

Divide $x$ by $1$ .

$x = \frac{4}{4}$

Simplify the right side.

Divide $4$ by $4$ .

$x = 1$

Step 5

Exclude the solutions that do not make $2 \sqrt{x} + 7 = 5$ true.

$No solution$

Solve for x 2 square root of x+7=5

Related Questions

Support LCMGCF.com by disabling your adblocker.