Question

7. Solve, using formula:
r2+x- (a + 2) (a + 1) = 0​

Answer 1

There are two proofs to this, and both are simple. One revolves around the Quadratic formula, while the other
involves writing ax2 + bx + c = a(x − r1)(x − r2) = ax2 − ax(r1 + r2) + ar1r2.
This allows us to find the sum and the product of the roots of any quadratic polynomial without actually
computing the roots themselves. (Sounds familiar?)
Example 1. Suppose p and q are the roots of the equation t
2 − 7t + 5. Find p
2 + q
2
.
Solution. Note that from our Vieta’s Formulas we have p + q = 7 and pq = 5. Therefore
p
2 + q
2 = (p + q)
2 − 2pq = 72 − 2 · 5 = 39 .

Example 2. Let m and n be the roots of the equation 2x
2 + 15x + 16 = 0. What is the value of 1
m
+
1
n
?
Solution. From Vieta’s Formulas, m + n = −
15
2
and mn =
16
2 = 8. Therefore
1
m
+
1
n
=
m + n
mn
=
−15/2
8
= −
15
16
.

3 The Cubic Case... and Beyond!
To see how Vieta’s Formulas can be expanded beyond quadratics, we look toward the cubic case for help. By using
a similar proof as we did in the previous section, we can write
x
3 + bx2 + cx + d = (x − r1)(x − r2)(x − r3)
= x
3 − (r1 + r2 + r3)x
2 + (r1r2 + r2r3 + r3r1)x − r1r2r3.
By compensating for the leading coefficient, we get another set of formulas:
Theorem 2. Let r1, r2, r3 be the roots of the cubic equation ax3 + bx2 + cx + d = 0. Then we have
r1 + r2 + r3 = −
b
a
, r1r2 + r2r3 + r3r1 =
c
a
, r1r2r3 = −
d
a
.
1Considering that the material in that session consists of basically stuff you should know, it should not make a difference on whether
or not you attended that lecture.