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

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

Solve for x

$$x=\sqrt{26}+4\approx 9.099019514$$

$$x=4-\sqrt{26}\approx -1.099019514$$

Steps Using the Quadratic Formula

Subtract \frac{1}{2}x^{2} from both sides.

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

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.

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

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

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

Square 4.

$$x=\frac{-4±\sqrt{16-4\left(-\frac{1}{2}\right)\times 5}}{2\left(-\frac{1}{2}\right)}$$

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

$$x=\frac{-4±\sqrt{16+2\times 5}}{2\left(-\frac{1}{2}\right)}$$

Multiply 2 times 5.

$$x=\frac{-4±\sqrt{16+10}}{2\left(-\frac{1}{2}\right)}$$

Add 16 to 10.

$$x=\frac{-4±\sqrt{26}}{2\left(-\frac{1}{2}\right)}$$

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

$$x=\frac{-4±\sqrt{26}}{-1}$$

Now solve the equation x=\frac{-4±\sqrt{26}}{-1} when ± is plus. Add -4 to \sqrt{26}.

$$x=\frac{\sqrt{26}-4}{-1}$$

Divide -4+\sqrt{26} by -1.

$$x=4-\sqrt{26}$$

Now solve the equation x=\frac{-4±\sqrt{26}}{-1} when ± is minus. Subtract \sqrt{26} from -4.

$$x=\frac{-\sqrt{26}-4}{-1}$$

Divide -4-\sqrt{26} by -1.

$$x=\sqrt{26}+4$$

The equation is now solved.

$$x=4-\sqrt{26} x=\sqrt{26}+4$$

Steps for Completing the Square

Subtract \frac{1}{2}x^{2} from both sides.

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

Subtract 5 from both sides. Anything subtracted from zero gives its negation.

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

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.

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

Multiply both sides by -2.

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

Dividing by -\frac{1}{2} undoes the multiplication by -\frac{1}{2}.

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

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

$$x^{2}-8x=-\frac{5}{-\frac{1}{2}}$$

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

$$x^{2}-8x=10$$

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

$$x^{2}-8x+\left(-4\right)^{2}=10+\left(-4\right)^{2}$$

Square -4.

$$x^{2}-8x+16=10+16$$

Add 10 to 16.

$$x^{2}-8x+16=26$$

Factor x^{2}-8x+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}.

$$\left(x-4\right)^{2}=26$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-4\right)^{2}}=\sqrt{26}$$

Simplify.

$$x-4=\sqrt{26} x-4=-\sqrt{26}$$

Add 4 to both sides of the equation.

$$x=\sqrt{26}+4 x=4-\sqrt{26}$$