Question

Part 1: Solve each of the quadratic equations below. Show your work. (3 points)

x2 – 36 = 0 and x2 = 8x – 12

Part 2: Describe what the solution(s) represent to the graph of each. (2 points)

Part 3: How are the graphs alike? How are they
different? (2 points)

Answer 1

So do you have to solve 2 question?

I'm talking about Part 1

Answer 2

x^2 – 36 = 0

First, replace the "b" variable by 0 since there isn't any. Then find the factors of -36 that adds up to 0.

x^2 + 0 - 36 = 0

1 * -36
2 * -18
3 * -12
4 * -9
6 * 6
-1 * 36
-2 * 18
-3 * 12
-4 * 9
-6 * 6

The one that adds up to 0 is 6 and -6 so now you factor normally by making them equal 0's.

(x + 6)(x - 6)
x+6 = 0
x-6 = 0

x= -6 and x= 6

Answer 3

I was actually answering that question on an online test a few moment ago what a coincidence

Answer 4

Unlike the others this one has x^2 in the beginning so you just subtract it from both sides. That way it'll equal 0 like a normal equation.

x^2 = 8x – 12
0 = -x^2 + 8x - 12

Because the -x^2 is negative we can't go on, but if we multiply both sides by -1 and now we have all positive numbers, then we can go on and find factors of 12 that add up to 8.

0 = x^2 + 8x + 12

1 * 12
2 * 6
3 * 4

2 and 6 will do, now we factor normally and make them
equal 0.

(x + 2)(x + 6)
x+2 = 0
x+6 = 0

x= -2 and x= -6

Answer 5

Do you get the steps?

Answer 6

@Rexy705 yeah I do.

Answer 7

what the solution x's equal tell u were the parabola will intersect the x-axis