By the quadratic formula, the roots are (-5 +- sqrt(69))/2, so the answer is ((-5 + sqrt(69)/2 + 3)((-5 - sqrt(69)/2 + 3) = -13.
Thank you so much, but I entered the answer after checking if it was correct, and it still says it was wrong...
Let a and b be the roots of \(y^2+5y-11=0\)
Find \((a+3)(b+3)\)
First
\(a+b=-5\)
\(ab=-11\)
\((a+3)(b+3)\) Expand this.
\(ab+3a+3b+9\) Factor 3 from middle terms
\(ab+3(a+b)+9\)
Now substitute the values
\(-11+3(-5)+9\)
\(-11-15+9\)
\(=-17\)
General note:
Vieta's formula states that:
for any quadratic equation (Even for any polynomial )
\(ax^2+bx+c=0\)
With roots: \((x_1,x_2)\)
\(x_1+x_2=-\frac{b}{a}\)
\(x_1x_2=\frac{c}{a}\)
Does it work for cubic equation?
\(ax^3+bx^2+cx+d=0\)
with roots: \((y_1,y_2,y_3)\)
\(y_1+y_2+y_3=-\frac{b}{a}\)
\(y_1y_2y_3=-\frac{d}{a}\)
and
\(y_1y_2*y_2y_3*y_1y_3=\frac{c}{a}\)
