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Let  and  be the roots of y^2 + 5y - 11 = 0. Find (a + 3)(b + 3).

 Apr 8, 2020
 #1
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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.

 Apr 8, 2020
 #2
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Thank you so much, but I entered the answer after checking if it was correct, and it still says it was wrong... 

 Apr 9, 2020
 #3
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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

\(y_1y_2*y_2y_3*y_1y_3=\frac{c}{a}\)

.
 Apr 9, 2020
 #5
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sorry!, y1+y2+y3=-d/a not b/a

Guest Apr 10, 2020
 #4
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Thank you so much! You saved me from a lifetime of shame xD

 Apr 9, 2020

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