Solve for x square root of 2x-1- square root of x+8=0

$\sqrt{2 x - 1} - \sqrt{x + 8} = 0$

Add $\sqrt{x + 8}$ to both sides of the equation.

$\sqrt{2 x - 1} = \sqrt{x + 8}$

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

${\sqrt{2 x - 1}}^{2} = {\sqrt{x + 8}}^{2}$

Simplify each side of the equation.

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

${({(2 x - 1)}^{\frac{1}{2}})}^{2} = {\sqrt{x + 8}}^{2}$

Simplify the left side.

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

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

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

${(2 x - 1)}^{\frac{1}{2} \cdot 2} = {\sqrt{x + 8}}^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(2 x - 1)}^{\frac{1}{2} \cdot 2} = {\sqrt{x + 8}}^{2}$

Rewrite the expression.

${(2 x - 1)}^{1} = {\sqrt{x + 8}}^{2}$

Simplify.

$2 x - 1 = {\sqrt{x + 8}}^{2}$

Simplify the right side.

Rewrite ${\sqrt{x + 8}}^{2}$ as $x + 8$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{x + 8}$ as ${(x + 8)}^{\frac{1}{2}}$ .

$2 x - 1 = {({(x + 8)}^{\frac{1}{2}})}^{2}$

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

$2 x - 1 = {(x + 8)}^{\frac{1}{2} \cdot 2}$

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

$2 x - 1 = {(x + 8)}^{\frac{2}{2}}$

Cancel the common factor of $2$ .

Cancel the common factor.

$2 x - 1 = {(x + 8)}^{\frac{2}{2}}$

Rewrite the expression.

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

Simplify.

$2 x - 1 = x + 8$

Solve for $x$ .

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

Subtract $x$ from both sides of the equation.

$2 x - 1 - x = 8$

Subtract $x$ from $2 x$ .

$x - 1 = 8$

Move all terms not containing $x$ to the right side of the equation.

Add $1$ to both sides of the equation.

$x = 8 + 1$

Add $8$ and $1$ .

$x = 9$

Solve for x square root of 2x-1- square root of x+8=0

Related Questions

Support LCMGCF.com by disabling your adblocker.