coolhabhi wrote:Let α and β be the roots of the equation x^2 - 3x - 3 = 0, then find the value of (β/α^2) + (α/β^2)
a)18
b)3
c)9
d)6
mevicks' solution is AWESOME, but as he stated, this question most likely goes beyond the scope of the GMAT
I just wanted to add a few things about how mevicks reached the conclusion that α + β =
3 and αβ =
-3
First recognize that we can't factor the expression x² - 3x - 3
Now, we COULD apply the quadratic formula to find the values of β and α, and then calculate (β/α²) + (α/β²), but the GMAT would never require us to make so many tedious and time-consuming calculations.
Instead, we might make a few observations about the roots of quadratic equations in the form x² - somethingx - something = 0
Some examples
x² - 2x - 15 = 0 is the same as (x - 5)(x + 3) = 0. So the roots are 5 and -3
Notice that 5 + (-3) =
2 and (5)(-3) =
15
x² - 5x - 6 = 0 is the same as (x - 6)(x + 1) = 0. So the roots are 6 and -1
Notice that 6 + (-1) =
5 and (6)(-1) =
-6
x² - 4x - 12 = 0 is the same as (x - 6)(x + 2) = 0. So the roots are 6 and -2
Notice that 6 + (-2) =
4 and (6)(-2) =
-12
So, if α and β are the roots of x² -
3x
- 3 = 0, then we know that α + β =
3 and αβ =
-3
We want to evaluate (β/α²) + (α/β²)
Notice that (β/α²) + (α/β²) = (β³/α²β²) + (α³/α²β²) = (β³+α³)/α²β²
Evaluate the denominator: α²β²
Since we already know that αβ =
-3, we can see that....
α²β² = (αβ)²
= (
-3)²
= 9
Evaluate the numerator: β³ + α³
THIS is where we have problems.
As mevicks pointed out, we need to recognize that β³ + α³ = (β + α)³ - 3βα(β + α)
This is a pretty ingenious observation, one that the GMAT would not require one to make under very severe time constraints. Likewise, the GMAT does not require us to know how to factor a sum of cubes this way [or the other way . . . a³ + b³ = (a + b)(a² – ab + b²)]
For these reasons, I believe this question is out of scope.
Cheers,
Brent
