Solve (4-3x)^2-64=0

$$( 4 - 3 x ) ^ { 2 } - 64 = 0$$

Solve for x

$$x = -\frac{4}{3} = -1\frac{1}{3} \approx -1.333333333$$

$$x=4$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(4-3x\right)^{2}.

$$16-24x+9x^{2}-64=0$$

Subtract 64 from 16 to get -48.

$$-48-24x+9x^{2}=0$$

Divide both sides by 3.

$$-16-8x+3x^{2}=0$$

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

$$3x^{2}-8x-16=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 3x^{2}+ax+bx-16. To find a and b, set up a system to be solved.

$$a+b=-8 ab=3\left(-16\right)=-48$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -48.

$$1,-48 2,-24 3,-16 4,-12 6,-8$$

Calculate the sum for each pair.

$$1-48=-47 2-24=-22 3-16=-13 4-12=-8 6-8=-2$$

The solution is the pair that gives sum -8.

$$a=-12 b=4$$

Rewrite 3x^{2}-8x-16 as \left(3x^{2}-12x\right)+\left(4x-16\right).

$$\left(3x^{2}-12x\right)+\left(4x-16\right)$$

Factor out 3x in the first and 4 in the second group.

$$3x\left(x-4\right)+4\left(x-4\right)$$

Factor out common term x-4 by using distributive property.

$$\left(x-4\right)\left(3x+4\right)$$

To find equation solutions, solve x-4=0 and 3x+4=0.

$$x=4 x=-\frac{4}{3}$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(4-3x\right)^{2}.

$$16-24x+9x^{2}-64=0$$

Subtract 64 from 16 to get -48.

$$-48-24x+9x^{2}=0$$

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.

$$9x^{2}-24x-48=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 9 for a, -24 for b, and -48 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-24\right)±\sqrt{\left(-24\right)^{2}-4\times 9\left(-48\right)}}{2\times 9}$$

Square -24.

$$x=\frac{-\left(-24\right)±\sqrt{576-4\times 9\left(-48\right)}}{2\times 9}$$

Multiply -4 times 9.

$$x=\frac{-\left(-24\right)±\sqrt{576-36\left(-48\right)}}{2\times 9}$$

Multiply -36 times -48.

$$x=\frac{-\left(-24\right)±\sqrt{576+1728}}{2\times 9}$$

Add 576 to 1728.

$$x=\frac{-\left(-24\right)±\sqrt{2304}}{2\times 9}$$

Take the square root of 2304.

$$x=\frac{-\left(-24\right)±48}{2\times 9}$$

The opposite of -24 is 24.

$$x=\frac{24±48}{2\times 9}$$

Multiply 2 times 9.

$$x=\frac{24±48}{18}$$

Now solve the equation x=\frac{24±48}{18} when ± is plus. Add 24 to 48.

$$x=\frac{72}{18}$$

Divide 72 by 18.

$$x=4$$

Now solve the equation x=\frac{24±48}{18} when ± is minus. Subtract 48 from 24.

$$x=-\frac{24}{18}$$

Reduce the fraction \frac{-24}{18} to lowest terms by extracting and canceling out 6.

$$x=-\frac{4}{3}$$

The equation is now solved.

$$x=4 x=-\frac{4}{3}$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(4-3x\right)^{2}.

$$16-24x+9x^{2}-64=0$$

Subtract 64 from 16 to get -48.

$$-48-24x+9x^{2}=0$$

Add 48 to both sides. Anything plus zero gives itself.

$$-24x+9x^{2}=48$$

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.

$$9x^{2}-24x=48$$

Divide both sides by 9.

$$\frac{9x^{2}-24x}{9}=\frac{48}{9}$$

Dividing by 9 undoes the multiplication by 9.

$$x^{2}+\left(-\frac{24}{9}\right)x=\frac{48}{9}$$

Reduce the fraction \frac{-24}{9} to lowest terms by extracting and canceling out 3.

$$x^{2}-\frac{8}{3}x=\frac{48}{9}$$

Reduce the fraction \frac{48}{9} to lowest terms by extracting and canceling out 3.

$$x^{2}-\frac{8}{3}x=\frac{16}{3}$$

Divide -\frac{8}{3}, the coefficient of the x term, by 2 to get -\frac{4}{3}. Then add the square of -\frac{4}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{8}{3}x+\left(-\frac{4}{3}\right)^{2}=\frac{16}{3}+\left(-\frac{4}{3}\right)^{2}$$

Square -\frac{4}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{8}{3}x+\frac{16}{9}=\frac{16}{3}+\frac{16}{9}$$

Add \frac{16}{3} to \frac{16}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{8}{3}x+\frac{16}{9}=\frac{64}{9}$$

Factor x^{2}-\frac{8}{3}x+\frac{16}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{4}{3}\right)^{2}=\frac{64}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{4}{3}\right)^{2}}=\sqrt{\frac{64}{9}}$$

Simplify.

$$x-\frac{4}{3}=\frac{8}{3} x-\frac{4}{3}=-\frac{8}{3}$$

Add \frac{4}{3} to both sides of the equation.

$$x=4 x=-\frac{4}{3}$$