Challenge question about quadratics?

[Sandstorm]

If the vertex of a quadratic equation is (6, -1) and the product of the roots is 35, what are the roots?

[Hawke]

Let a and b represent the two roots to the quadratic equation. Then, since the x-coordinate of the vertex is the average of the two zeroes, we have:

(a + b)/2 = 6
==> a + b = 12. . . . . . . . . . . . . . .(1)

We are then given that the product of the roots is 35. So:

ab = 35. . . . . . . . . . . . . . . . . . . ..(2)

Solving (1) for b gives:

b = 12 - a. . . . . . . . . . . . . . . . . . .(3)

Substituting (3) into (2) yields:

a(12 - a) = 35
==> 12a - a^2 = 35
==> a^2 - 12a + 35 = 0
==> (a - 5)(a - 7) = 0
==> a = 5 and a = 7.

Then:
i) a = 5 ==> b = 12 - a = 12 - 5 = 7
ii) a = 7 ==> b = 12 - a = 12 - 7 = 5.
(Either way, the roots are 5 and 7)

Therefore, the roots are 5 and 7.

I hope this helps!