Question

2. Using Vièta's theorem, find the solutions to the equation. a) x^2 - 3x + 2 = 0 b) x^2 + 2x - 15 = 0.

Answer 1

Given:

$\begin{gathered} x^2-3x+2=0 \\ x^2+2x-15=0 \end{gathered}$

Required:

We need to find the solution by Vièta's theorem.

Explanation:

Compare 1st equation with

$ax^2+bx+c=0$

we get

$\begin{gathered} a=1 \\ b=-3 \\ c=2 \end{gathered}$

Vièta's theorem is

$\begin{gathered} x_1+x_2=-\frac{b}{a} \\ x_1x_2=\frac{c}{a} \end{gathered}$

$\begin{gathered} x_1+x_2=3 \\ x_1x_2=2 \end{gathered}$

now solve this equation and we get

$\begin{gathered} x_1=1 \\ x_2=2 \end{gathered}$

because addition of 1 and 2 is 3 and multiplication is 2

Now for 2nd equation

$\begin{gathered} a=1 \\ b=2 \\ c=-15 \end{gathered}$

apply Vièta's theorem

$\begin{gathered} x_1+x_2=-2 \\ x_1x_2=-15 \end{gathered}$

by this

$\begin{gathered} x_1=3 \\ x_2=-5 \end{gathered}$

because addition of 3 and -5 is -2 and multiplication is -15