Segment AB contains point C. AC=3x^2+4x+9 CB=-x^2+6x+2 AB= 23cm. Find BC.

Answer 1

If AB contains C, then AC + CB = AC.

First, we need to solve for x, so that we can ultimately plug that into the formula for CB = -x^2 + 6x + 2.

Since AC + CB = AC, we can add to the two formulas for each of these and combine like terms, and set them equal to the total distance of AB = 23cm.

Add 3x^2 + 4x + 9 and -x^2 + 6x + 2, and get 2x^2 + 10x + 11 = 23.

Subtract 23 to get 2x^2 + 10x - 12 = 0.

From there, we can use the quadratic formula, or x = -b ± √(b^2 - 4ac) / 2a, where Ax^2 + BX + C = 0

Therefore, A=2, B=10, C=-12

When we plug those numbers into the quadratic formula, we get:

-10 ± √(10^2 - 4(2)(-12)) / 2(2)

-10 ± √(100 + 96) / 4

-10 ± √(196) / 4

-10 ± 14 / 4

Once we've simplified as much as possible, we must separate the plus/minus and solve each.

-10 + 14 / 4

4/4 = 1

And:

-10 - 14 / 4

-24/4 = -6

Then, we can plug x = -6 and x = 1 into -x^2 + 6x + 2.

First, x = -6

-(-6)^2 + 6(-6) + 2

-36 -36 + 2 = -70 -> It can't be this answer because we can't have a negative distance. Therefore..

x = 1

-(-1)^2 + 6(1) + 2

-1 + 6 + 2 = 7

BC = 7cm

Answer 2

3x² + 4x + 9 + x² + 6x + 2 = 23

Can you solve for x and answer?