Solve 1/2a+3aquada=5

$$\frac { 1 } { 2 } a + 3 a \quad a = 5$$

Solve for a

$$a = \frac{\sqrt{241} - 1}{12} \approx 1.210347891$$

$$a=\frac{-\sqrt{241}-1}{12}\approx -1.377014558$$

Steps Using the Quadratic Formula

Multiply a and a to get a^{2}.

$$\frac{1}{2}a+3a^{2}=5$$

Subtract 5 from both sides.

$$\frac{1}{2}a+3a^{2}-5=0$$

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.

$$3a^{2}+\frac{1}{2}a-5=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, \frac{1}{2} for b, and -5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$a=\frac{-\frac{1}{2}±\sqrt{\left(\frac{1}{2}\right)^{2}-4\times 3\left(-5\right)}}{2\times 3}$$

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

$$a=\frac{-\frac{1}{2}±\sqrt{\frac{1}{4}-4\times 3\left(-5\right)}}{2\times 3}$$

Multiply -4 times 3.

$$a=\frac{-\frac{1}{2}±\sqrt{\frac{1}{4}-12\left(-5\right)}}{2\times 3}$$

Multiply -12 times -5.

$$a=\frac{-\frac{1}{2}±\sqrt{\frac{1}{4}+60}}{2\times 3}$$

Add \frac{1}{4} to 60.

$$a=\frac{-\frac{1}{2}±\sqrt{\frac{241}{4}}}{2\times 3}$$

Take the square root of \frac{241}{4}.

$$a=\frac{-\frac{1}{2}±\frac{\sqrt{241}}{2}}{2\times 3}$$

Multiply 2 times 3.

$$a=\frac{-\frac{1}{2}±\frac{\sqrt{241}}{2}}{6}$$

Now solve the equation a=\frac{-\frac{1}{2}±\frac{\sqrt{241}}{2}}{6} when ± is plus. Add -\frac{1}{2} to \frac{\sqrt{241}}{2}.

$$a=\frac{\sqrt{241}-1}{2\times 6}$$

Divide \frac{-1+\sqrt{241}}{2} by 6.

$$a=\frac{\sqrt{241}-1}{12}$$

Now solve the equation a=\frac{-\frac{1}{2}±\frac{\sqrt{241}}{2}}{6} when ± is minus. Subtract \frac{\sqrt{241}}{2} from -\frac{1}{2}.

$$a=\frac{-\sqrt{241}-1}{2\times 6}$$

Divide \frac{-1-\sqrt{241}}{2} by 6.

$$a=\frac{-\sqrt{241}-1}{12}$$

The equation is now solved.

$$a=\frac{\sqrt{241}-1}{12} a=\frac{-\sqrt{241}-1}{12}$$

Steps for Completing the Square

Multiply a and a to get a^{2}.

$$\frac{1}{2}a+3a^{2}=5$$

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.

$$3a^{2}+\frac{1}{2}a=5$$

Divide both sides by 3.

$$\frac{3a^{2}+\frac{1}{2}a}{3}=\frac{5}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$a^{2}+\frac{\frac{1}{2}}{3}a=\frac{5}{3}$$

Divide \frac{1}{2} by 3.

$$a^{2}+\frac{1}{6}a=\frac{5}{3}$$

Divide \frac{1}{6}, the coefficient of the x term, by 2 to get \frac{1}{12}. Then add the square of \frac{1}{12} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$a^{2}+\frac{1}{6}a+\left(\frac{1}{12}\right)^{2}=\frac{5}{3}+\left(\frac{1}{12}\right)^{2}$$

Square \frac{1}{12} by squaring both the numerator and the denominator of the fraction.

$$a^{2}+\frac{1}{6}a+\frac{1}{144}=\frac{5}{3}+\frac{1}{144}$$

Add \frac{5}{3} to \frac{1}{144} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$a^{2}+\frac{1}{6}a+\frac{1}{144}=\frac{241}{144}$$

Factor a^{2}+\frac{1}{6}a+\frac{1}{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}.

$$\left(a+\frac{1}{12}\right)^{2}=\frac{241}{144}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(a+\frac{1}{12}\right)^{2}}=\sqrt{\frac{241}{144}}$$

Simplify.

$$a+\frac{1}{12}=\frac{\sqrt{241}}{12} a+\frac{1}{12}=-\frac{\sqrt{241}}{12}$$

Subtract \frac{1}{12} from both sides of the equation.

$$a=\frac{\sqrt{241}-1}{12} a=\frac{-\sqrt{241}-1}{12}$$