June 23 - Vieta's Formulas

June 23 - Vieta's Formulas

Roman Hall

Vieta's Formulas for Quadratics

Let  $r_1$  and  $r_2$  be the roots of the quadratic equation  $ax^2+bx+c=0$ . Then the two identities  $r_1+r_2=-\frac{b}{a}$  and  $r_1r_2=\frac{c}{a}$  both hold.

Open Ended

1. Suppose  $p$  and  $q$  are the roots of the equation  $t^2-7t+5=0$ . Compute  $p^2+q^2$ .

Solution to Problem 1

Note that from our Vieta's Formulas we have  $p+q=7$  and  $pq=5$ . Therefore  $p^2+q^2=\left(p+q\right)^2-2pq=7^2-2\cdot5=39$ . 

Open Ended

2. Let  $m$  and  $n$  be the roots of the equation  $2x^2+15x+16=0$ . Compute  $\frac{1}{m}+\frac{1}{n}$ .

Solution to Problem 2

From Vieta's Formulas,  $m+n=-\frac{15}{2}$  and  $mn=\frac{16}{2}=8$ . 

Therefore  $\frac{1}{m}+\frac{1}{n}=\frac{m+n}{mn}=\frac{-\frac{15}{2}}{8}=\ -\frac{15}{16}$ .

Vieta's Formulas for Cubics

Let  $r_1,r_2,r_3$  be the roots of the cubic equation  $ax^3+bx^2+cx+d=0$ . Then we have

 $r_1+r_2+r_3=-\frac{b}{a}$ ,  $r_1r_2+r_1r_3+r_2r_3=\frac{c}{a}$ ,  $r_1r_2r_3=-\frac{d}{a}$ .

Open Ended

3. Suppose  $p$ ,  $q$ , and  $r$  are the roots of the polynomial  $t^3-2t^3+3t-4$ . Compute  $\left(p+1\right)\left(q+1\right)\left(r+1\right)$ .

Solution to Problem 3

 $\left(p+1\right)\left(q+1\right)\left(r+1\right)=pqr+\left(pq+pr+qr\right)+\left(p+q+r\right)+1$ 

 $4+3+2+1=10$  

Open Ended

4. The roots  $r_1$ ,  $r_2$ , and  $r_3$  of  $x^3-2x^2-11x+a$  satisfy  $r_1+2r_2+3r_3=0$ . Compute all possible values of  $a$ .

Solution to Problem 4

From Vieta's Formulas, we have  $r_1+r_2+r_3=2$ ,  $r_1r_2+r_1r_3+r_2r_3=-11$ ,  $r_1+2r_2+3r_3=0$ . Subtracting the first equation from the third gives  $r_2=-2-2r_3$ . Plugging that into the first equation gives  $r_1=r_3+4$ . Substituting all that into the second equation gives  $\left(r_3+4\right)\left(-2-2r_3\right)+\left(r_3+4\right)r_3+\left(-2-2r_3\right)r_3=-11$ . Simplifying and solving gives  $r_3=\frac{1}{3},-3$ , so  $\left(r_1,r_2,r_3\right)=\left(\frac{13}{3},-\frac{8}{3},\frac{1}{3}\right),\left(1,4,-3\right)$ . It follows that  $a=-r_1r_2r_3=\frac{104}{27},12$ .