Solve x^2/4+5x=-1/2

$$\frac { x ^ { 2 } } { 4 } + 5 x = \frac { - 1 } { 2 }$$

Solve for x

$$x=7\sqrt{2}-10\approx -0.100505063$$

$$x=-7\sqrt{2}-10\approx -19.899494937$$

Steps Using the Quadratic Formula

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$\frac{1}{4}x^{2}+5x=-\frac{1}{2}$$

Add \frac{1}{2} to both sides of the equation.

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

Subtracting -\frac{1}{2} from itself leaves 0.

$$\frac{1}{4}x^{2}+5x-\left(-\frac{1}{2}\right)=0$$

Subtract -\frac{1}{2} from 0.

$$\frac{1}{4}x^{2}+5x+\frac{1}{2}=0$$

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

$$x=\frac{-5±\sqrt{5^{2}-4\times \frac{1}{4}\times \frac{1}{2}}}{2\times \frac{1}{4}}$$

Square 5.

$$x=\frac{-5±\sqrt{25-4\times \frac{1}{4}\times \frac{1}{2}}}{2\times \frac{1}{4}}$$

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

$$x=\frac{-5±\sqrt{25-\frac{1}{2}}}{2\times \frac{1}{4}}$$

Add 25 to -\frac{1}{2}.

$$x=\frac{-5±\sqrt{\frac{49}{2}}}{2\times \frac{1}{4}}$$

Take the square root of \frac{49}{2}.

$$x=\frac{-5±\frac{7\sqrt{2}}{2}}{2\times \frac{1}{4}}$$

Multiply 2 times \frac{1}{4}.

$$x=\frac{-5±\frac{7\sqrt{2}}{2}}{\frac{1}{2}}$$

Now solve the equation x=\frac{-5±\frac{7\sqrt{2}}{2}}{\frac{1}{2}} when ± is plus. Add -5 to \frac{7\sqrt{2}}{2}.

$$x=\frac{\frac{7\sqrt{2}}{2}-5}{\frac{1}{2}}$$

Divide -5+\frac{7\sqrt{2}}{2} by \frac{1}{2} by multiplying -5+\frac{7\sqrt{2}}{2} by the reciprocal of \frac{1}{2}.

$$x=7\sqrt{2}-10$$

Now solve the equation x=\frac{-5±\frac{7\sqrt{2}}{2}}{\frac{1}{2}} when ± is minus. Subtract \frac{7\sqrt{2}}{2} from -5.

$$x=\frac{-\frac{7\sqrt{2}}{2}-5}{\frac{1}{2}}$$

Divide -5-\frac{7\sqrt{2}}{2} by \frac{1}{2} by multiplying -5-\frac{7\sqrt{2}}{2} by the reciprocal of \frac{1}{2}.

$$x=-7\sqrt{2}-10$$

The equation is now solved.

$$x=7\sqrt{2}-10 x=-7\sqrt{2}-10$$

Steps for Completing the Square

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$\frac{1}{4}x^{2}+5x=-\frac{1}{2}$$

Multiply both sides by 4.

$$\frac{\frac{1}{4}x^{2}+5x}{\frac{1}{4}}=-\frac{\frac{1}{2}}{\frac{1}{4}}$$

Dividing by \frac{1}{4} undoes the multiplication by \frac{1}{4}.

$$x^{2}+\frac{5}{\frac{1}{4}}x=-\frac{\frac{1}{2}}{\frac{1}{4}}$$

Divide 5 by \frac{1}{4} by multiplying 5 by the reciprocal of \frac{1}{4}.

$$x^{2}+20x=-\frac{\frac{1}{2}}{\frac{1}{4}}$$

Divide -\frac{1}{2} by \frac{1}{4} by multiplying -\frac{1}{2} by the reciprocal of \frac{1}{4}.

$$x^{2}+20x=-2$$

Divide 20, the coefficient of the x term, by 2 to get 10. Then add the square of 10 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+20x+10^{2}=-2+10^{2}$$

Square 10.

$$x^{2}+20x+100=-2+100$$

Add -2 to 100.

$$x^{2}+20x+100=98$$

Factor x^{2}+20x+100. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+10\right)^{2}=98$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+10\right)^{2}}=\sqrt{98}$$

Simplify.

$$x+10=7\sqrt{2} x+10=-7\sqrt{2}$$

Subtract 10 from both sides of the equation.

$$x=7\sqrt{2}-10 x=-7\sqrt{2}-10$$