Solve Using the Quadratic Formula 3/4k^2-k-8/3=0

$\frac{3}{4} k^{2} - k - \frac{8}{3} = 0$

Step 1

Simplify the left side.

Combine $\frac{3}{4}$ and $k^{2}$ .

$\frac{3 k^{2}}{4} - k - \frac{8}{3} = 0$

Step 2

Multiply through by the least common denominator $12$ , then simplify.

Apply the distributive property.

$12 (\frac{3 k^{2}}{4}) + 12 (- k) + 12 (- \frac{8}{3}) = 0$

Simplify.

Cancel the common factor of $4$ .

Factor $4$ out of $12$ .

$4 (3) (\frac{3 k^{2}}{4}) + 12 (- k) + 12 (- \frac{8}{3}) = 0$

Cancel the common factor.

$4 \cdot (3 (\frac{3 k^{2}}{4})) + 12 (- k) + 12 (- \frac{8}{3}) = 0$

Rewrite the expression.

$3 (3 k^{2}) + 12 (- k) + 12 (- \frac{8}{3}) = 0$

Multiply $3$ by $3$ .

$9 k^{2} + 12 (- k) + 12 (- \frac{8}{3}) = 0$

Multiply $- 1$ by $12$ .

$9 k^{2} - 12 k + 12 (- \frac{8}{3}) = 0$

Cancel the common factor of $3$ .

Move the leading negative in $- \frac{8}{3}$ into the numerator.

$9 k^{2} - 12 k + 12 (\frac{- 8}{3}) = 0$

Factor $3$ out of $12$ .

$9 k^{2} - 12 k + 3 (4) (\frac{- 8}{3}) = 0$

Cancel the common factor.

$9 k^{2} - 12 k + 3 \cdot (4 (\frac{- 8}{3})) = 0$

Rewrite the expression.

$9 k^{2} - 12 k + 4 \cdot - 8 = 0$

Multiply $4$ by $- 8$ .

$9 k^{2} - 12 k - 32 = 0$

Step 3

Use the quadratic formula to find the solutions.

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

Step 4

Substitute the values $a = 9$ , $b = - 12$ , and $c = - 32$ into the quadratic formula and solve for $k$ .

$\frac{12 \pm \sqrt{{(- 12)}^{2} - 4 \cdot (9 \cdot - 32)}}{2 \cdot 9}$

Step 5

Simplify.

Simplify the numerator.

Raise $- 12$ to the power of $2$ .

$k = \frac{12 \pm \sqrt{144 - 4 \cdot 9 \cdot - 32}}{2 \cdot 9}$

Multiply $- 4 \cdot 9 \cdot - 32$ .

Multiply $- 4$ by $9$ .

$k = \frac{12 \pm \sqrt{144 - 36 \cdot - 32}}{2 \cdot 9}$

Multiply $- 36$ by $- 32$ .

$k = \frac{12 \pm \sqrt{144 + 1152}}{2 \cdot 9}$

Add $144$ and $1152$ .

$k = \frac{12 \pm \sqrt{1296}}{2 \cdot 9}$

Rewrite $1296$ as $36^{2}$ .

$k = \frac{12 \pm \sqrt{36^{2}}}{2 \cdot 9}$

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

$k = \frac{12 \pm 36}{2 \cdot 9}$

Multiply $2$ by $9$ .

$k = \frac{12 \pm 36}{18}$

Simplify $\frac{12 \pm 36}{18}$ .

$k = \frac{2 \pm 6}{3}$

Step 6

The final answer is the combination of both solutions.

$k = \frac{8}{3}, - \frac{4}{3}$

Solve Using the Quadratic Formula 3/4k^2-k-8/3=0

Related Questions

Support LCMGCF.com by disabling your adblocker.