Solve left(3x-5right)^2=0

$${ \left(3x-5 \right) }^{ 2 } =0$$

Solve for x

$$x = \frac{5}{3} = 1\frac{2}{3} \approx 1.666666667$$

Steps Using Factoring By Grouping

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

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 9x^{2}+ax+bx+25. To find a and b, set up a system to be solved.

$$a+b=-30 ab=9\times 25=225$$

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 225.

$$-1,-225 -3,-75 -5,-45 -9,-25 -15,-15$$

Calculate the sum for each pair.

$$-1-225=-226 -3-75=-78 -5-45=-50 -9-25=-34 -15-15=-30$$

The solution is the pair that gives sum -30.

$$a=-15 b=-15$$

Rewrite 9x^{2}-30x+25 as \left(9x^{2}-15x\right)+\left(-15x+25\right).

$$\left(9x^{2}-15x\right)+\left(-15x+25\right)$$

Factor out 3x in the first and -5 in the second group.

$$3x\left(3x-5\right)-5\left(3x-5\right)$$

Factor out common term 3x-5 by using distributive property.

$$\left(3x-5\right)\left(3x-5\right)$$

Rewrite as a binomial square.

$$\left(3x-5\right)^{2}$$

To find equation solution, solve 3x-5=0.

$$x=\frac{5}{3}$$

Steps Using the Quadratic Formula

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

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 9 for a, -30 for b, and 25 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-30\right)±\sqrt{\left(-30\right)^{2}-4\times 9\times 25}}{2\times 9}$$

Square -30.

$$x=\frac{-\left(-30\right)±\sqrt{900-4\times 9\times 25}}{2\times 9}$$

Multiply -4 times 9.

$$x=\frac{-\left(-30\right)±\sqrt{900-36\times 25}}{2\times 9}$$

Multiply -36 times 25.

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

Add 900 to -900.

$$x=\frac{-\left(-30\right)±\sqrt{0}}{2\times 9}$$

Take the square root of 0.

$$x=-\frac{-30}{2\times 9}$$

The opposite of -30 is 30.

$$x=\frac{30}{2\times 9}$$

Multiply 2 times 9.

$$x=\frac{30}{18}$$

Reduce the fraction \frac{30}{18} to lowest terms by extracting and canceling out 6.

$$x=\frac{5}{3}$$

Steps for Completing the Square

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

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

Subtract 25 from both sides. Anything subtracted from zero gives its negation.

$$9x^{2}-30x=-25$$

Divide both sides by 9.

$$\frac{9x^{2}-30x}{9}=-\frac{25}{9}$$

Dividing by 9 undoes the multiplication by 9.

$$x^{2}+\left(-\frac{30}{9}\right)x=-\frac{25}{9}$$

Reduce the fraction \frac{-30}{9} to lowest terms by extracting and canceling out 3.

$$x^{2}-\frac{10}{3}x=-\frac{25}{9}$$

Divide -\frac{10}{3}, the coefficient of the x term, by 2 to get -\frac{5}{3}. Then add the square of -\frac{5}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{10}{3}x+\left(-\frac{5}{3}\right)^{2}=-\frac{25}{9}+\left(-\frac{5}{3}\right)^{2}$$

Square -\frac{5}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{10}{3}x+\frac{25}{9}=\frac{-25+25}{9}$$

Add -\frac{25}{9} to \frac{25}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{10}{3}x+\frac{25}{9}=0$$

Factor x^{2}-\frac{10}{3}x+\frac{25}{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}.

$$\left(x-\frac{5}{3}\right)^{2}=0$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{3}\right)^{2}}=\sqrt{0}$$

Simplify.

$$x-\frac{5}{3}=0 x-\frac{5}{3}=0$$

Add \frac{5}{3} to both sides of the equation.

$$x=\frac{5}{3} x=\frac{5}{3}$$

The equation is now solved. Solutions are the same.

$$x=\frac{5}{3}$$