# Math Help - Just 3 problems I can't figure out guys!

1. ## Just 3 problems I can't figure out guys!

Could you please help me solve these 3? The total assignment is 16 questions but these are the 3 troubling me.

1)

x^2/3 + 3x^1/3 - 18 = 0

Solution set is ( , )

2)

(1/y-9)^2 + 4(1/y-9) - 21 = 0

3)

a= x^2 + 16x -7

( , )

2. Originally Posted by knowledgekick
1) x^2/3 + 3x^1/3 - 18 = 0
You can convert this to quadratic "form":

. . . . . $\left(x^{\frac{1}{3}}\right)^2\, +\, 3\left(x^{\frac{1}{3}}\right)\, -\, 18\, =\, 0$

Factor this quadratic, and then solve the two exponential equations which result.

Originally Posted by knowledgekick
2) (1/y-9)^2 + 4(1/y-9) - 21 = 0
This is the same sort of thing, except that the resulting equations will be rational equations.

. . . . . $\left[\left(\frac{1}{y\, -\, 9}\right)\, +\, 7\right]\left[\left(\frac{1}{y\, -\, 9}\right)\, -\, 3\right]\, =\, 0$

Originally Posted by knowledgekick
3) a= x^2 + 16x -7
All you can do with this one, since you don't have a value for "a", is to move the unknown over to the right-hand side:

. . . . . $0\, =\, x^2\, +\, 16x\, -\, 7\, -\, a$

Then apply the Quadratic Formula, with the Formula's a = 1, b = 16, and c = -7 - a. (The two a's are different, obviously.)

If you get stuck, please reply showing how far you have gotten. Thank you!