Solve 4a^2-9/2a-3=9

$$\frac { 4 a ^ { 2 } - 9 } { 2 a - 3 } = 9$$

Solve for a

$$a=3$$

Steps Using Factoring By Grouping

Variable a cannot be equal to \frac{3}{2} since division by zero is not defined. Multiply both sides of the equation by 2a-3.

$$4a^{2}-9=9\left(2a-3\right)$$

Use the distributive property to multiply 9 by 2a-3.

$$4a^{2}-9=18a-27$$

Subtract 18a from both sides.

$$4a^{2}-9-18a=-27$$

Add 27 to both sides.

$$4a^{2}-9-18a+27=0$$

Add -9 and 27 to get 18.

$$4a^{2}+18-18a=0$$

Divide both sides by 2.

$$2a^{2}+9-9a=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$2a^{2}-9a+9=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 2a^{2}+aa+ba+9. To find a and b, set up a system to be solved.

$$a+b=-9 ab=2\times 9=18$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 18.

$$-1,-18 -2,-9 -3,-6$$

Calculate the sum for each pair.

$$-1-18=-19 -2-9=-11 -3-6=-9$$

The solution is the pair that gives sum -9.

$$a=-6 b=-3$$

Rewrite 2a^{2}-9a+9 as \left(2a^{2}-6a\right)+\left(-3a+9\right).

$$\left(2a^{2}-6a\right)+\left(-3a+9\right)$$

Factor out 2a in the first and -3 in the second group.

$$2a\left(a-3\right)-3\left(a-3\right)$$

Factor out common term a-3 by using distributive property.

$$\left(a-3\right)\left(2a-3\right)$$

To find equation solutions, solve a-3=0 and 2a-3=0.

$$a=3 a=\frac{3}{2}$$

Variable a cannot be equal to \frac{3}{2}.

$$a=3$$

Steps Using the Quadratic Formula

Variable a cannot be equal to \frac{3}{2} since division by zero is not defined. Multiply both sides of the equation by 2a-3.

$$4a^{2}-9=9\left(2a-3\right)$$

Use the distributive property to multiply 9 by 2a-3.

$$4a^{2}-9=18a-27$$

Subtract 18a from both sides.

$$4a^{2}-9-18a=-27$$

Add 27 to both sides.

$$4a^{2}-9-18a+27=0$$

Add -9 and 27 to get 18.

$$4a^{2}+18-18a=0$$

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.

$$4a^{2}-18a+18=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -18 for b, and 18 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$a=\frac{-\left(-18\right)±\sqrt{\left(-18\right)^{2}-4\times 4\times 18}}{2\times 4}$$

Square -18.

$$a=\frac{-\left(-18\right)±\sqrt{324-4\times 4\times 18}}{2\times 4}$$

Multiply -4 times 4.

$$a=\frac{-\left(-18\right)±\sqrt{324-16\times 18}}{2\times 4}$$

Multiply -16 times 18.

$$a=\frac{-\left(-18\right)±\sqrt{324-288}}{2\times 4}$$

Add 324 to -288.

$$a=\frac{-\left(-18\right)±\sqrt{36}}{2\times 4}$$

Take the square root of 36.

$$a=\frac{-\left(-18\right)±6}{2\times 4}$$

The opposite of -18 is 18.

$$a=\frac{18±6}{2\times 4}$$

Multiply 2 times 4.

$$a=\frac{18±6}{8}$$

Now solve the equation a=\frac{18±6}{8} when ± is plus. Add 18 to 6.

$$a=\frac{24}{8}$$

Divide 24 by 8.

$$a=3$$

Now solve the equation a=\frac{18±6}{8} when ± is minus. Subtract 6 from 18.

$$a=\frac{12}{8}$$

Reduce the fraction \frac{12}{8} to lowest terms by extracting and canceling out 4.

$$a=\frac{3}{2}$$

The equation is now solved.

$$a=3 a=\frac{3}{2}$$

Variable a cannot be equal to \frac{3}{2}.

$$a=3$$

Steps for Completing the Square

Variable a cannot be equal to \frac{3}{2} since division by zero is not defined. Multiply both sides of the equation by 2a-3.

$$4a^{2}-9=9\left(2a-3\right)$$

Use the distributive property to multiply 9 by 2a-3.

$$4a^{2}-9=18a-27$$

Subtract 18a from both sides.

$$4a^{2}-9-18a=-27$$

Add 9 to both sides.

$$4a^{2}-18a=-27+9$$

Add -27 and 9 to get -18.

$$4a^{2}-18a=-18$$

Divide both sides by 4.

$$\frac{4a^{2}-18a}{4}=-\frac{18}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$a^{2}+\left(-\frac{18}{4}\right)a=-\frac{18}{4}$$

Reduce the fraction \frac{-18}{4} to lowest terms by extracting and canceling out 2.

$$a^{2}-\frac{9}{2}a=-\frac{18}{4}$$

Reduce the fraction \frac{-18}{4} to lowest terms by extracting and canceling out 2.

$$a^{2}-\frac{9}{2}a=-\frac{9}{2}$$

Divide -\frac{9}{2}, the coefficient of the x term, by 2 to get -\frac{9}{4}. Then add the square of -\frac{9}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$a^{2}-\frac{9}{2}a+\left(-\frac{9}{4}\right)^{2}=-\frac{9}{2}+\left(-\frac{9}{4}\right)^{2}$$

Square -\frac{9}{4} by squaring both the numerator and the denominator of the fraction.

$$a^{2}-\frac{9}{2}a+\frac{81}{16}=-\frac{9}{2}+\frac{81}{16}$$

Add -\frac{9}{2} to \frac{81}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$a^{2}-\frac{9}{2}a+\frac{81}{16}=\frac{9}{16}$$

Factor a^{2}-\frac{9}{2}a+\frac{81}{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}.

$$\left(a-\frac{9}{4}\right)^{2}=\frac{9}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(a-\frac{9}{4}\right)^{2}}=\sqrt{\frac{9}{16}}$$

Simplify.

$$a-\frac{9}{4}=\frac{3}{4} a-\frac{9}{4}=-\frac{3}{4}$$

Add \frac{9}{4} to both sides of the equation.

$$a=3 a=\frac{3}{2}$$

Variable a cannot be equal to \frac{3}{2}.

$$a=3$$