Solve (5z-3)^2-9z^2=0

$$( 5 z - 3 ) ^ { 2 } - 9 z ^ { 2 } = 0$$

Solve for z

$$z=\frac{3}{8}=0.375$$

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

Steps Using Factoring By Grouping

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(5z-3\right)^{2}.

$$25z^{2}-30z+9-9z^{2}=0$$

Combine 25z^{2} and -9z^{2} to get 16z^{2}.

$$16z^{2}-30z+9=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 16z^{2}+az+bz+9. To find a and b, set up a system to be solved.

$$a+b=-30 ab=16\times 9=144$$

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 144.

$$-1,-144 -2,-72 -3,-48 -4,-36 -6,-24 -8,-18 -9,-16 -12,-12$$

Calculate the sum for each pair.

$$-1-144=-145 -2-72=-74 -3-48=-51 -4-36=-40 -6-24=-30 -8-18=-26 -9-16=-25 -12-12=-24$$

The solution is the pair that gives sum -30.

$$a=-24 b=-6$$

Rewrite 16z^{2}-30z+9 as \left(16z^{2}-24z\right)+\left(-6z+9\right).

$$\left(16z^{2}-24z\right)+\left(-6z+9\right)$$

Factor out 8z in the first and -3 in the second group.

$$8z\left(2z-3\right)-3\left(2z-3\right)$$

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

$$\left(2z-3\right)\left(8z-3\right)$$

To find equation solutions, solve 2z-3=0 and 8z-3=0.

$$z=\frac{3}{2} z=\frac{3}{8}$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(5z-3\right)^{2}.

$$25z^{2}-30z+9-9z^{2}=0$$

Combine 25z^{2} and -9z^{2} to get 16z^{2}.

$$16z^{2}-30z+9=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 16 for a, -30 for b, and 9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$z=\frac{-\left(-30\right)±\sqrt{\left(-30\right)^{2}-4\times 16\times 9}}{2\times 16}$$

Square -30.

$$z=\frac{-\left(-30\right)±\sqrt{900-4\times 16\times 9}}{2\times 16}$$

Multiply -4 times 16.

$$z=\frac{-\left(-30\right)±\sqrt{900-64\times 9}}{2\times 16}$$

Multiply -64 times 9.

$$z=\frac{-\left(-30\right)±\sqrt{900-576}}{2\times 16}$$

Add 900 to -576.

$$z=\frac{-\left(-30\right)±\sqrt{324}}{2\times 16}$$

Take the square root of 324.

$$z=\frac{-\left(-30\right)±18}{2\times 16}$$

The opposite of -30 is 30.

$$z=\frac{30±18}{2\times 16}$$

Multiply 2 times 16.

$$z=\frac{30±18}{32}$$

Now solve the equation z=\frac{30±18}{32} when ± is plus. Add 30 to 18.

$$z=\frac{48}{32}$$

Reduce the fraction \frac{48}{32} to lowest terms by extracting and canceling out 16.

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

Now solve the equation z=\frac{30±18}{32} when ± is minus. Subtract 18 from 30.

$$z=\frac{12}{32}$$

Reduce the fraction \frac{12}{32} to lowest terms by extracting and canceling out 4.

$$z=\frac{3}{8}$$

The equation is now solved.

$$z=\frac{3}{2} z=\frac{3}{8}$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(5z-3\right)^{2}.

$$25z^{2}-30z+9-9z^{2}=0$$

Combine 25z^{2} and -9z^{2} to get 16z^{2}.

$$16z^{2}-30z+9=0$$

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

$$16z^{2}-30z=-9$$

Divide both sides by 16.

$$\frac{16z^{2}-30z}{16}=-\frac{9}{16}$$

Dividing by 16 undoes the multiplication by 16.

$$z^{2}+\left(-\frac{30}{16}\right)z=-\frac{9}{16}$$

Reduce the fraction \frac{-30}{16} to lowest terms by extracting and canceling out 2.

$$z^{2}-\frac{15}{8}z=-\frac{9}{16}$$

Divide -\frac{15}{8}, the coefficient of the x term, by 2 to get -\frac{15}{16}. Then add the square of -\frac{15}{16} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$z^{2}-\frac{15}{8}z+\left(-\frac{15}{16}\right)^{2}=-\frac{9}{16}+\left(-\frac{15}{16}\right)^{2}$$

Square -\frac{15}{16} by squaring both the numerator and the denominator of the fraction.

$$z^{2}-\frac{15}{8}z+\frac{225}{256}=-\frac{9}{16}+\frac{225}{256}$$

Add -\frac{9}{16} to \frac{225}{256} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$z^{2}-\frac{15}{8}z+\frac{225}{256}=\frac{81}{256}$$

Factor z^{2}-\frac{15}{8}z+\frac{225}{256}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(z-\frac{15}{16}\right)^{2}=\frac{81}{256}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(z-\frac{15}{16}\right)^{2}}=\sqrt{\frac{81}{256}}$$

Simplify.

$$z-\frac{15}{16}=\frac{9}{16} z-\frac{15}{16}=-\frac{9}{16}$$

Add \frac{15}{16} to both sides of the equation.

$$z=\frac{3}{2} z=\frac{3}{8}$$