Solve for k square root of 2k^2+17-7=0

$\sqrt{2 k^{2} + 17} - 7 = 0$

Step 1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{2 k^{2} + 17}$ as ${(2 k^{2} + 17)}^{\frac{1}{2}}$ .

${(2 k^{2} + 17)}^{\frac{1}{2}} - 7 = 0$

Step 2

Add $7$ to both sides of the equation.

${(2 k^{2} + 17)}^{\frac{1}{2}} = 7$

Step 3

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

${({(2 k^{2} + 17)}^{\frac{1}{2}})}^{2} = 7^{2}$

Step 4

Simplify the exponent.

Simplify the left side.

Simplify ${({(2 k^{2} + 17)}^{\frac{1}{2}})}^{2}$ .

Multiply the exponents in ${({(2 k^{2} + 17)}^{\frac{1}{2}})}^{2}$ .

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

${(2 k^{2} + 17)}^{\frac{1}{2} \cdot 2} = 7^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(2 k^{2} + 17)}^{\frac{1}{2} \cdot 2} = 7^{2}$

Rewrite the expression.

${(2 k^{2} + 17)}^{1} = 7^{2}$

Simplify.

$2 k^{2} + 17 = 7^{2}$

Simplify the right side.

Raise $7$ to the power of $2$ .

$2 k^{2} + 17 = 49$

Step 5

Solve for $k$ .

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

Subtract $17$ from both sides of the equation.

$2 k^{2} = 49 - 17$

Subtract $17$ from $49$ .

$2 k^{2} = 32$

Divide each term in $2 k^{2} = 32$ by $2$ and simplify.

Divide each term in $2 k^{2} = 32$ by $2$ .

$\frac{2 k^{2}}{2} = \frac{32}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 k^{2}}{2} = \frac{32}{2}$

Divide $k^{2}$ by $1$ .

$k^{2} = \frac{32}{2}$

Simplify the right side.

Divide $32$ by $2$ .

$k^{2} = 16$

Take the square root of both sides of the equation to eliminate the exponent on the left side.

$k = \pm \sqrt{16}$

Simplify $\pm \sqrt{16}$ .

Rewrite $16$ as $4^{2}$ .

$k = \pm \sqrt{4^{2}}$

Pull terms out from under the radical, assuming positive real numbers.

$k = \pm 4$

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$k = 4$

Next, use the negative value of the $\pm$ to find the second solution.

$k = - 4$

The complete solution is the result of both the positive and negative portions of the solution.

$k = 4, - 4$

Solve for k square root of 2k^2+17-7=0

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