Solve (k+1/4)^2-1/16-1/5=0

$$( k + \frac { 1 } { 4 } ) ^ { 2 } - \frac { 1 } { 16 } - \frac { 1 } { 5 } = 0$$

Solve for k

$$k=\frac{\sqrt{105}}{20}-\frac{1}{4}\approx 0.262347538$$

$$k=-\frac{\sqrt{105}}{20}-\frac{1}{4}\approx -0.762347538$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(k+\frac{1}{4}\right)^{2}.

$$k^{2}+\frac{1}{2}k+\frac{1}{16}-\frac{1}{16}-\frac{1}{5}=0$$

Subtract \frac{1}{16} from \frac{1}{16} to get 0.

$$k^{2}+\frac{1}{2}k-\frac{1}{5}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, \frac{1}{2} for b, and -\frac{1}{5} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$k=\frac{-\frac{1}{2}±\sqrt{\left(\frac{1}{2}\right)^{2}-4\left(-\frac{1}{5}\right)}}{2}$$

Square \frac{1}{2} by squaring both the numerator and the denominator of the fraction.

$$k=\frac{-\frac{1}{2}±\sqrt{\frac{1}{4}-4\left(-\frac{1}{5}\right)}}{2}$$

Multiply -4 times -\frac{1}{5}.

$$k=\frac{-\frac{1}{2}±\sqrt{\frac{1}{4}+\frac{4}{5}}}{2}$$

Add \frac{1}{4} to \frac{4}{5} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$k=\frac{-\frac{1}{2}±\sqrt{\frac{21}{20}}}{2}$$

Take the square root of \frac{21}{20}.

$$k=\frac{-\frac{1}{2}±\frac{\sqrt{105}}{10}}{2}$$

Now solve the equation k=\frac{-\frac{1}{2}±\frac{\sqrt{105}}{10}}{2} when ± is plus. Add -\frac{1}{2} to \frac{\sqrt{105}}{10}.

$$k=\frac{\frac{\sqrt{105}}{10}-\frac{1}{2}}{2}$$

Divide -\frac{1}{2}+\frac{\sqrt{105}}{10} by 2.

$$k=\frac{\sqrt{105}}{20}-\frac{1}{4}$$

Now solve the equation k=\frac{-\frac{1}{2}±\frac{\sqrt{105}}{10}}{2} when ± is minus. Subtract \frac{\sqrt{105}}{10} from -\frac{1}{2}.

$$k=\frac{-\frac{\sqrt{105}}{10}-\frac{1}{2}}{2}$$

Divide -\frac{1}{2}-\frac{\sqrt{105}}{10} by 2.

$$k=-\frac{\sqrt{105}}{20}-\frac{1}{4}$$

The equation is now solved.

$$k=\frac{\sqrt{105}}{20}-\frac{1}{4} k=-\frac{\sqrt{105}}{20}-\frac{1}{4}$$

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(k+\frac{1}{4}\right)^{2}.

$$k^{2}+\frac{1}{2}k+\frac{1}{16}-\frac{1}{16}-\frac{1}{5}=0$$

Subtract \frac{1}{16} from \frac{1}{16} to get 0.

$$k^{2}+\frac{1}{2}k-\frac{1}{5}=0$$

Add \frac{1}{5} to both sides. Anything plus zero gives itself.

$$k^{2}+\frac{1}{2}k=\frac{1}{5}$$

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

$$k^{2}+\frac{1}{2}k+\left(\frac{1}{4}\right)^{2}=\frac{1}{5}+\left(\frac{1}{4}\right)^{2}$$

Square \frac{1}{4} by squaring both the numerator and the denominator of the fraction.

$$k^{2}+\frac{1}{2}k+\frac{1}{16}=\frac{1}{5}+\frac{1}{16}$$

Add \frac{1}{5} to \frac{1}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$k^{2}+\frac{1}{2}k+\frac{1}{16}=\frac{21}{80}$$

Factor k^{2}+\frac{1}{2}k+\frac{1}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(k+\frac{1}{4}\right)^{2}=\frac{21}{80}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(k+\frac{1}{4}\right)^{2}}=\sqrt{\frac{21}{80}}$$

Simplify.

$$k+\frac{1}{4}=\frac{\sqrt{105}}{20} k+\frac{1}{4}=-\frac{\sqrt{105}}{20}$$

Subtract \frac{1}{4} from both sides of the equation.

$$k=\frac{\sqrt{105}}{20}-\frac{1}{4} k=-\frac{\sqrt{105}}{20}-\frac{1}{4}$$