Solve this algrebra problem?

Question

t^2/3 + t^1/3 - 6 = 0

Please tell how you got the answer too because I need to understand it for a test coming up. Thanks so much.

Answer 1

t = 8
This is solved by writing

(t^(1/3))² + (t^(1/3)) - 6 = 0 then letting x = t^(1/3) so we get

x² + x - 6 = 0 Now use the quadratic formula, solve for x, then substitute back for t.

Answer 2

By letting x = t^(1/3) you can rewrite the equation as

x^2 + x - 6 = 0 since x^2 = [t^(1/3]^2 = t^2/3

Now factor to get

(x+3)(x-2)=0 which has roots x = -3, 2 by equating each

factor =0 and solving.

But you are not looking for x, you want t.

So go back to your substitution x = t^1/3 and set x=-3 and 2

-3 = t^1/3 and 2 = t^1/3 If you raise each side to the

3rd power you will get

-27 = t and 8 = t since [t^1/3]^3 = t^3/3 = t

Answer 3

T^(2/3+1/3)=t^(1)=t
t-6=0
t=6

Answer 4

t^2/3 + t^1/3 - 6 = 0

we can write this as a quadratic equation
let x= t^(1/3) to get



x^2 +x -6 =0

Discriminant = 1+24 =25

x1 = (-1-5)/2 =-3
x2= (-1+5)/2 =2

so
t^(1/3) =-3
no solution

t^(1/3) =2

t = 2^3 =8

verification
8^(2/3) -8^(1/3) -6 =0

Answer 5

Let y = t^(1/3)
y ² + y - 6 = 0
( y + 3 ) ( y - 2 ) = 0
y = - 3 , y = 2
t^(1/3) = - 3 , t^(1/3) = 2
t = - 27 , t = 8

Answer 6

(t^1/3 + 3)(t^1/3 -2) = 0, so

t^1/3 = -3, or t^1/3 = 2, so

t = -27, or t = 8.

Hope this helps, Twiggy.