Solve 4k^2-4k-4=-5k-5

$$4 k ^ { 2 } - 4 k - 4 = - 5 k - 5$$

Solve for k

$$k=\frac{-1+\sqrt{15}i}{8}\approx -0.125+0.484122918i$$

$$k=\frac{-\sqrt{15}i-1}{8}\approx -0.125-0.484122918i$$

Steps Using the Quadratic Formula

Add 5k to both sides.

$$4k^{2}-4k-4+5k=-5$$

Combine -4k and 5k to get k.

$$4k^{2}+k-4=-5$$

Add 5 to both sides.

$$4k^{2}+k-4+5=0$$

Add -4 and 5 to get 1.

$$4k^{2}+k+1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, 1 for b, and 1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$k=\frac{-1±\sqrt{1^{2}-4\times 4}}{2\times 4}$$

Square 1.

$$k=\frac{-1±\sqrt{1-4\times 4}}{2\times 4}$$

Multiply -4 times 4.

$$k=\frac{-1±\sqrt{1-16}}{2\times 4}$$

Add 1 to -16.

$$k=\frac{-1±\sqrt{-15}}{2\times 4}$$

Take the square root of -15.

$$k=\frac{-1±\sqrt{15}i}{2\times 4}$$

Multiply 2 times 4.

$$k=\frac{-1±\sqrt{15}i}{8}$$

Now solve the equation k=\frac{-1±\sqrt{15}i}{8} when ± is plus. Add -1 to i\sqrt{15}.

$$k=\frac{-1+\sqrt{15}i}{8}$$

Now solve the equation k=\frac{-1±\sqrt{15}i}{8} when ± is minus. Subtract i\sqrt{15} from -1.

$$k=\frac{-\sqrt{15}i-1}{8}$$

The equation is now solved.

$$k=\frac{-1+\sqrt{15}i}{8} k=\frac{-\sqrt{15}i-1}{8}$$

Steps for Completing the Square

Add 5k to both sides.

$$4k^{2}-4k-4+5k=-5$$

Combine -4k and 5k to get k.

$$4k^{2}+k-4=-5$$

Add 4 to both sides.

$$4k^{2}+k=-5+4$$

Add -5 and 4 to get -1.

$$4k^{2}+k=-1$$

Divide both sides by 4.

$$\frac{4k^{2}+k}{4}=-\frac{1}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$k^{2}+\frac{1}{4}k=-\frac{1}{4}$$

Divide \frac{1}{4}, the coefficient of the x term, by 2 to get \frac{1}{8}. Then add the square of \frac{1}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$k^{2}+\frac{1}{4}k+\left(\frac{1}{8}\right)^{2}=-\frac{1}{4}+\left(\frac{1}{8}\right)^{2}$$

Square \frac{1}{8} by squaring both the numerator and the denominator of the fraction.

$$k^{2}+\frac{1}{4}k+\frac{1}{64}=-\frac{1}{4}+\frac{1}{64}$$

Add -\frac{1}{4} to \frac{1}{64} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$k^{2}+\frac{1}{4}k+\frac{1}{64}=-\frac{15}{64}$$

Factor k^{2}+\frac{1}{4}k+\frac{1}{64}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(k+\frac{1}{8}\right)^{2}=-\frac{15}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(k+\frac{1}{8}\right)^{2}}=\sqrt{-\frac{15}{64}}$$

Simplify.

$$k+\frac{1}{8}=\frac{\sqrt{15}i}{8} k+\frac{1}{8}=-\frac{\sqrt{15}i}{8}$$

Subtract \frac{1}{8} from both sides of the equation.

$$k=\frac{-1+\sqrt{15}i}{8} k=\frac{-\sqrt{15}i-1}{8}$$