Respuesta :
Answer:
The value of [tex]r_{1}^2+r_{2}^2[/tex] is 126
Step-by-step explanation:
We can use the definition of Vieta's formula for quadratics:
Given [tex]f(x) = ax^{2} +bx+c[/tex], if the equation f(x) = 0 has roots [tex]r_{1}[/tex] and [tex]r_{2}[/tex] then
[tex]r_{1}+r_{2}=-\frac{b}{a} , r_{1}\cdot r_{2}=\frac{c}{a}[/tex]
So suppose [tex]r_{1}[/tex] and [tex]r_{2}[/tex] are the roots of the equation [tex]x^{2} - 12x + 9[/tex] to find [tex]r_{1}^2+r_{2}^2[/tex], note that from our Vieta's formula for quadratics we have
[tex]r_{1}+r_{2}=-\frac{-12}{1}\\r_{1}+r_{2}=12[/tex] and [tex]r_{1}\cdot r_{2}=\frac{9}{1}\\r_{1}\cdot r_{2}=9[/tex]
Therefore
[tex]r_{1}^2+r_{2}^2=(r_{1}+r_{2})^2-2\cdot r_{1}r_{2}\\r_{1}^2+r_{2}^2=(12)^2-2\cdot 9\\r_{1}^2+r_{2}^2=126[/tex]
