Question

solve the equation for all values of x by completing the square.
\\(3x^2 + 60 = 36x\\)

Explanation:

Step1: Rearrange the equation

First, we want to get the equation in the form \(ax^{2}+bx = c\). Subtract \(36x\) from both sides of the equation \(3x^{2}+60 = 36x\) to get:
\(3x^{2}-36x + 60=0\)
Then, divide the entire equation by \(3\) to simplify:
\(x^{2}-12x + 20 = 0\)
Now, we have the equation in the form \(x^{2}+bx = c\) (after moving the constant to the other side). Subtract \(20\) from both sides:
\(x^{2}-12x=- 20\)

Step2: Complete the square

To complete the square for the quadratic expression \(x^{2}-12x\), we take half of the coefficient of \(x\), square it, and add it to both sides. The coefficient of \(x\) is \(- 12\), half of it is \(\frac{-12}{2}=-6\), and squaring it gives \((-6)^{2} = 36\). Add \(36\) to both sides of the equation \(x^{2}-12x=-20\):
\(x^{2}-12x + 36=-20 + 36\)
The left - hand side is now a perfect square trinomial, which factors as \((x - 6)^{2}\). So we have:
\((x - 6)^{2}=16\)

Step3: Solve for x

Take the square root of both sides. Remember that when we take the square root of a number, we get both a positive and a negative solution:
\(x - 6=\pm\sqrt{16}=\pm4\)
Now, solve for \(x\) in two cases:

• Case 1: \(x-6 = 4\), then \(x=4 + 6=10\)
• Case 2: \(x - 6=-4\), then \(x=-4 + 6 = 2\)

Answer:

\(x = 2\) or \(x = 10\)