Question

19 solve the equation $-3x^2 + 36x + 6 = 0$ by completing the square and fill in the boxes below with the exact solution values. (for example, if your solutions turned out to be $-37 pm sqrt{71}$, you would type -37 in the left box and 71 in the right box) $x = square pm sqrt{square}$

Explanation:

Step1: Divide by -3

Divide the entire equation \(-3x^{2}+36x + 6 = 0\) by \(-3\) to make the coefficient of \(x^{2}\) equal to 1.
We get \(x^{2}-12x - 2=0\). Then, rewrite it as \(x^{2}-12x=2\).

Step2: Complete the square

To complete the square for the \(x\) terms, take half of the coefficient of \(x\) (which is \(- 12\)), square it. Half of \(-12\) is \(-6\), and squaring \(-6\) gives \(36\).
Add \(36\) to both sides of the equation \(x^{2}-12x = 2\):
\(x^{2}-12x + 36=2 + 36\)
The left - hand side is a perfect square trinomial, \((x - 6)^{2}=38\)

Step3: Solve for x

Take the square root of both sides: \(x-6=\pm\sqrt{38}\)
Then, add 6 to both sides to solve for \(x\): \(x = 6\pm\sqrt{38}\)

Answer:

The left box (the number before the \(\pm\)) is \(6\) and the right box (the number inside the square root) is \(38\). So the answers are \(6\) (left box) and \(38\) (right box).