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# uhh, help again? sorry

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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

$$y_1y_2*y_2y_3*y_1y_3=\frac{c}{a}$$

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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