Solve (a-3)^2=0

$${( a - 3 )} ^ { 2 } = 0$$

Solve for a

$$a=3$$

Steps Using Factoring

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

$$a^{2}-6a+9=0$$

To solve the equation, factor a^{2}-6a+9 using formula a^{2}+\left(a+b\right)a+ab=\left(a+a\right)\left(a+b\right). To find a and b, set up a system to be solved.

$$a+b=-6 ab=9$$

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 9.

$$-1,-9 -3,-3$$

Calculate the sum for each pair.

$$-1-9=-10 -3-3=-6$$

The solution is the pair that gives sum -6.

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

Rewrite factored expression \left(a+a\right)\left(a+b\right) using the obtained values.

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

Rewrite as a binomial square.

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

To find equation solution, solve a-3=0.

$$a=3$$

Steps Using Factoring By Grouping

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

$$a^{2}-6a+9=0$$

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

$$a+b=-6 ab=1\times 9=9$$

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 9.

$$-1,-9 -3,-3$$

Calculate the sum for each pair.

$$-1-9=-10 -3-3=-6$$

The solution is the pair that gives sum -6.

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

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

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

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

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

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

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

Rewrite as a binomial square.

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

To find equation solution, solve a-3=0.

$$a=3$$

Steps Using the Quadratic Formula

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

$$a^{2}-6a+9=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -6 for b, and 9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$a=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 9}}{2}$$

Square -6.

$$a=\frac{-\left(-6\right)±\sqrt{36-4\times 9}}{2}$$

Multiply -4 times 9.

$$a=\frac{-\left(-6\right)±\sqrt{36-36}}{2}$$

Add 36 to -36.

$$a=\frac{-\left(-6\right)±\sqrt{0}}{2}$$

Take the square root of 0.

$$a=-\frac{-6}{2}$$

The opposite of -6 is 6.

$$a=\frac{6}{2}$$

Divide 6 by 2.

$$a=3$$

Steps for Completing the Square

Take the square root of both sides of the equation.

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

Simplify.

$$a-3=0 a-3=0$$

Add 3 to both sides of the equation.

$$a=3 a=3$$

The equation is now solved. Solutions are the same.

$$a=3$$