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Sub 505 (Easy)|   Algebra|      
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grantcke
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If –6 and 3 are the solutions to the equation x^2 + bx + c = 0, where Updated on: 02 Nov 2016, 02:08
Originally posted by grantcke on 02 Nov 2016, 02:01.
Last edited by Bunuel on 02 Nov 2016, 02:08, edited 1 time in total.
Edited the question.
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If –6 and 3 are the solutions to the equation x^2 + bx + c = 0, where b and c are constants, what is the value of b + c?

A. –18
B. –15
C. –3
D. 3
E. 8
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B
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Bunuel
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If –6 and 3 are the solutions to the equation x^2 + bx + c = 0, where 02 Nov 2016, 02:13
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grantcke
If –6 and 3 are the solutions to the equation x^2 + bx + c = 0, where b and c are constants, what is the value of b + c?

A. –18
B. –15
C. –3
D. 3
E. 8
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Viete's theorem states that for the roots \(x_1\) and \(x_2\) of a quadratic equation \(ax^2+bx+c=0\):

\(x_1+x_2=\frac{-b}{a}\) AND \(x_1*x_2=\frac{c}{a}\).

Thus for x^2 + bx + c = 0:

-6 + 3 = -b/1 --> b = 3 and (-6)*3 = c/1 --> c = -18.

b + c = 3 - 18 = -15.

Answer: B.

Hope it's clear.
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grantcke
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If –6 and 3 are the solutions to the equation x^2 + bx + c = 0, where 02 Nov 2016, 02:44
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Hi Bunuel,

Thanks thats a great way to work it out. I made two equations in the end and solved for each variable independently.

Bunuel
grantcke
If –6 and 3 are the solutions to the equation x^2 + bx + c = 0, where b and c are constants, what is the value of b + c?

A. –18
B. –15
C. –3
D. 3
E. 8

Viete's theorem states that for the roots \(x_1\) and \(x_2\) of a quadratic equation \(ax^2+bx+c=0\):

\(x_1+x_2=\frac{-b}{a}\) AND \(x_1*x_2=\frac{c}{a}\).

Thus for x^2 + bx + c = 0:

-6 + 3 = -b/1 --> b = 3 and (-6)*3 = c/1 --> c = -18.

b + c = 3 - 18 = -15.

Answer: B.

Hope it's clear.
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If –6 and 3 are the solutions to the equation x^2 + bx + c = 0, where 02 Nov 2016, 03:02
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–6 and 3 are the solutions to the equation x^2 + bx + c = 0
= x^2 + bx + c = (x+6)(x-3) -> x^2+3x-18
we get b=3 and c=-18 ->b+c=3+(-18)=-15
B.
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If –6 and 3 are the solutions to the equation x^2 + bx + c = 0, where 21 May 2018, 16:42
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grantcke
If –6 and 3 are the solutions to the equation x^2 + bx + c = 0, where b and c are constants, what is the value of b + c?

A. –18
B. –15
C. –3
D. 3
E. 8
Show more

Since the zeroes are x = -6 or x = 3, we know we can factor the expression x^2 + bx + c as (x + 6)(x - 3), and we can create the equation:

(x + 6)(x - 3) = x^2 + bx + c

x^2 + 3x - 18 = x^2 + bx + c

Thus, we see that b = 3 and c = -18 for a sum of -15.

Answer: B
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If –6 and 3 are the solutions to the equation x^2 + bx + c = 0, where 12 Dec 2018, 10:52
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according to vete's theorem ,

x1+x2=-b/a , and x1*x2= c/a

which gives -6+3=-b--> b=3

-3*6=c--> c=18
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If –6 and 3 are the solutions to the equation x^2 + bx + c = 0, where 10 Jan 2024, 01:27
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