Algebra Examples

$k^{2} - 31 - 2 k = - 6 - 3 k^{2} - 2 k$

Step 1

Move all terms to the left side of the equation and simplify.

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Step 1.1

Move all the expressions to the left side of the equation.

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Step 1.1.1

Add $6$ to both sides of the equation.

$k^{2} - 31 - 2 k + 6 = - 3 k^{2} - 2 k$

Step 1.1.2

Add $3 k^{2}$ to both sides of the equation.

$k^{2} - 31 - 2 k + 6 + 3 k^{2} = - 2 k$

Step 1.1.3

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

$k^{2} - 31 - 2 k + 6 + 3 k^{2} + 2 k = 0$

Step 1.2

Simplify $k^{2} - 31 - 2 k + 6 + 3 k^{2} + 2 k$ .

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Step 1.2.1

Combine the opposite terms in $k^{2} - 31 - 2 k + 6 + 3 k^{2} + 2 k$ .

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Step 1.2.1.1

Add $- 2 k$ and $2 k$ .

$k^{2} - 31 + 6 + 3 k^{2} + 0 = 0$

Step 1.2.1.2

Add $k^{2} - 31 + 6 + 3 k^{2}$ and $0$ .

$k^{2} - 31 + 6 + 3 k^{2} = 0$

Step 1.2.2

Add $k^{2}$ and $3 k^{2}$ .

$4 k^{2} - 31 + 6 = 0$

Step 1.2.3

Add $- 31$ and $6$ .

$4 k^{2} - 25 = 0$

Step 2

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 3

Substitute the values $a = 4$ , $b = 0$ , and $c = - 25$ into the quadratic formula and solve for $k$ .

$\frac{0 \pm \sqrt{0^{2} - 4 \cdot (4 \cdot - 25)}}{2 \cdot 4}$

Step 4

Simplify.

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Step 4.1

Simplify the numerator.

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Step 4.1.1

Raising $0$ to any positive power yields $0$ .

$k = \frac{0 \pm \sqrt{0 - 4 \cdot 4 \cdot - 25}}{2 \cdot 4}$

Step 4.1.2

Multiply $- 4 \cdot 4 \cdot - 25$ .

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Step 4.1.2.1

Multiply $- 4$ by $4$ .

$k = \frac{0 \pm \sqrt{0 - 16 \cdot - 25}}{2 \cdot 4}$

Step 4.1.2.2

Multiply $- 16$ by $- 25$ .

$k = \frac{0 \pm \sqrt{0 + 400}}{2 \cdot 4}$

Step 4.1.3

Add $0$ and $400$ .

$k = \frac{0 \pm \sqrt{400}}{2 \cdot 4}$

Step 4.1.4

Rewrite $400$ as $20^{2}$ .

$k = \frac{0 \pm \sqrt{20^{2}}}{2 \cdot 4}$

Step 4.1.5

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

$k = \frac{0 \pm 20}{2 \cdot 4}$

Step 4.2

Multiply $2$ by $4$ .

$k = \frac{0 \pm 20}{8}$

Step 4.3

Simplify $\frac{0 \pm 20}{8}$ .

$k = \frac{\pm 5}{2}$

Step 5

The final answer is the combination of both solutions.

$k = \frac{5}{2}, - \frac{5}{2}$