Solve 6k^2=-49k-8

$$6 k ^ { 2 } = - 49 k - 8$$

Solve for k

$$k=-8$$

$$k=-\frac{1}{6}\approx -0.166666667$$

Steps Using Factoring By Grouping

Add 49k to both sides.

$$6k^{2}+49k=-8$$

Add 8 to both sides.

$$6k^{2}+49k+8=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 6k^{2}+ak+bk+8. To find a and b, set up a system to be solved.

$$a+b=49 ab=6\times 8=48$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 48.

$$1,48 2,24 3,16 4,12 6,8$$

Calculate the sum for each pair.

$$1+48=49 2+24=26 3+16=19 4+12=16 6+8=14$$

The solution is the pair that gives sum 49.

$$a=1 b=48$$

Rewrite 6k^{2}+49k+8 as \left(6k^{2}+k\right)+\left(48k+8\right).

$$\left(6k^{2}+k\right)+\left(48k+8\right)$$

Factor out k in the first and 8 in the second group.

$$k\left(6k+1\right)+8\left(6k+1\right)$$

Factor out common term 6k+1 by using distributive property.

$$\left(6k+1\right)\left(k+8\right)$$

To find equation solutions, solve 6k+1=0 and k+8=0.

$$k=-\frac{1}{6} k=-8$$

Steps Using the Quadratic Formula

Add 49k to both sides.

$$6k^{2}+49k=-8$$

Add 8 to both sides.

$$6k^{2}+49k+8=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 6 for a, 49 for b, and 8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$k=\frac{-49±\sqrt{49^{2}-4\times 6\times 8}}{2\times 6}$$

Square 49.

$$k=\frac{-49±\sqrt{2401-4\times 6\times 8}}{2\times 6}$$

Multiply -4 times 6.

$$k=\frac{-49±\sqrt{2401-24\times 8}}{2\times 6}$$

Multiply -24 times 8.

$$k=\frac{-49±\sqrt{2401-192}}{2\times 6}$$

Add 2401 to -192.

$$k=\frac{-49±\sqrt{2209}}{2\times 6}$$

Take the square root of 2209.

$$k=\frac{-49±47}{2\times 6}$$

Multiply 2 times 6.

$$k=\frac{-49±47}{12}$$

Now solve the equation k=\frac{-49±47}{12} when ± is plus. Add -49 to 47.

$$k=-\frac{2}{12}$$

Reduce the fraction \frac{-2}{12} to lowest terms by extracting and canceling out 2.

$$k=-\frac{1}{6}$$

Now solve the equation k=\frac{-49±47}{12} when ± is minus. Subtract 47 from -49.

$$k=-\frac{96}{12}$$

Divide -96 by 12.

$$k=-8$$

The equation is now solved.

$$k=-\frac{1}{6} k=-8$$

Steps for Completing the Square

Add 49k to both sides.

$$6k^{2}+49k=-8$$

Divide both sides by 6.

$$\frac{6k^{2}+49k}{6}=-\frac{8}{6}$$

Dividing by 6 undoes the multiplication by 6.

$$k^{2}+\frac{49}{6}k=-\frac{8}{6}$$

Reduce the fraction \frac{-8}{6} to lowest terms by extracting and canceling out 2.

$$k^{2}+\frac{49}{6}k=-\frac{4}{3}$$

Divide \frac{49}{6}, the coefficient of the x term, by 2 to get \frac{49}{12}. Then add the square of \frac{49}{12} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$k^{2}+\frac{49}{6}k+\left(\frac{49}{12}\right)^{2}=-\frac{4}{3}+\left(\frac{49}{12}\right)^{2}$$

Square \frac{49}{12} by squaring both the numerator and the denominator of the fraction.

$$k^{2}+\frac{49}{6}k+\frac{2401}{144}=-\frac{4}{3}+\frac{2401}{144}$$

Add -\frac{4}{3} to \frac{2401}{144} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$k^{2}+\frac{49}{6}k+\frac{2401}{144}=\frac{2209}{144}$$

Factor k^{2}+\frac{49}{6}k+\frac{2401}{144}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(k+\frac{49}{12}\right)^{2}=\frac{2209}{144}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(k+\frac{49}{12}\right)^{2}}=\sqrt{\frac{2209}{144}}$$

Simplify.

$$k+\frac{49}{12}=\frac{47}{12} k+\frac{49}{12}=-\frac{47}{12}$$

Subtract \frac{49}{12} from both sides of the equation.

$$k=-\frac{1}{6} k=-8$$