Solve 4x^2+9=12x

$$4 x ^ { 2 } + 9 = 12 x$$

Solve for x

$$x = \frac{3}{2} = 1\frac{1}{2} = 1.5$$

Steps Using Factoring By Grouping

Subtract 12x from both sides.

$$4x^{2}+9-12x=0$$

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

$$4x^{2}-12x+9=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 4x^{2}+ax+bx+9. To find a and b, set up a system to be solved.

$$a+b=-12 ab=4\times 9=36$$

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 36.

$$-1,-36 -2,-18 -3,-12 -4,-9 -6,-6$$

Calculate the sum for each pair.

$$-1-36=-37 -2-18=-20 -3-12=-15 -4-9=-13 -6-6=-12$$

The solution is the pair that gives sum -12.

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

Rewrite 4x^{2}-12x+9 as \left(4x^{2}-6x\right)+\left(-6x+9\right).

$$\left(4x^{2}-6x\right)+\left(-6x+9\right)$$

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

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

Factor out common term 2x-3 by using distributive property.

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

Rewrite as a binomial square.

$$\left(2x-3\right)^{2}$$

To find equation solution, solve 2x-3=0.

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

Steps Using the Quadratic Formula

Subtract 12x from both sides.

$$4x^{2}+9-12x=0$$

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.

$$4x^{2}-12x+9=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -12 for b, and 9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 4\times 9}}{2\times 4}$$

Square -12.

$$x=\frac{-\left(-12\right)±\sqrt{144-4\times 4\times 9}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-\left(-12\right)±\sqrt{144-16\times 9}}{2\times 4}$$

Multiply -16 times 9.

$$x=\frac{-\left(-12\right)±\sqrt{144-144}}{2\times 4}$$

Add 144 to -144.

$$x=\frac{-\left(-12\right)±\sqrt{0}}{2\times 4}$$

Take the square root of 0.

$$x=-\frac{-12}{2\times 4}$$

The opposite of -12 is 12.

$$x=\frac{12}{2\times 4}$$

Multiply 2 times 4.

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

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

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

Steps for Completing the Square

Subtract 12x from both sides.

$$4x^{2}+9-12x=0$$

Subtract 9 from both sides. Anything subtracted from zero gives its negation.

$$4x^{2}-12x=-9$$

Divide both sides by 4.

$$\frac{4x^{2}-12x}{4}=-\frac{9}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}+\left(-\frac{12}{4}\right)x=-\frac{9}{4}$$

Divide -12 by 4.

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

Divide -3, the coefficient of the x term, by 2 to get -\frac{3}{2}. Then add the square of -\frac{3}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-3x+\left(-\frac{3}{2}\right)^{2}=-\frac{9}{4}+\left(-\frac{3}{2}\right)^{2}$$

Square -\frac{3}{2} by squaring both the numerator and the denominator of the fraction.

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

Add -\frac{9}{4} to \frac{9}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-3x+\frac{9}{4}=0$$

Factor x^{2}-3x+\frac{9}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{3}{2}\right)^{2}=0$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{2}\right)^{2}}=\sqrt{0}$$

Simplify.

$$x-\frac{3}{2}=0 x-\frac{3}{2}=0$$

Add \frac{3}{2} to both sides of the equation.

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

The equation is now solved. Solutions are the same.

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