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
The OA is B.
I'm confused with this PS question. Please, can any expert assist me with it? Thanks in advanced.
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If -6 and 3 are the solutions of the...
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BTGmoderatorLU
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Given x² + bx + c = 0:
-b = the SUM of the two solutions.
c = the PRODUCT of the two solutions.
-b = -6 + 3
-b = -3
b = 3.
c = the PRODUCT of the two solutions:
c = -6*3
c = -18.
Thus:
b+c = 3 + (-18) = -15.
The correct answer is B.
-b = the SUM of the two solutions.
c = the PRODUCT of the two solutions.
-b = the SUM of the two solutions:LUANDATO wrote: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
-b = -6 + 3
-b = -3
b = 3.
c = the PRODUCT of the two solutions:
c = -6*3
c = -18.
Thus:
b+c = 3 + (-18) = -15.
The correct answer is B.
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I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
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Matt@VeritasPrep
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Another way:
Remember that the roots of (x + r) * (x + s) = 0 are x = -r and x = -s.
We're told that -r = -6 and -s = 3, so r = 6 and s = -3. Plugging those in, we have
(x + 6) * (x + -3) = 0
and foiling
x*x + 6x - 3x - 18 = 0
or
x*x + 3x - 18 = 0
so b = 3, c = -18, and b + c = -15.
Remember that the roots of (x + r) * (x + s) = 0 are x = -r and x = -s.
We're told that -r = -6 and -s = 3, so r = 6 and s = -3. Plugging those in, we have
(x + 6) * (x + -3) = 0
and foiling
x*x + 6x - 3x - 18 = 0
or
x*x + 3x - 18 = 0
so b = 3, c = -18, and b + c = -15.
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Matt@VeritasPrep
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Yet another way:
If our solutions are -6 and 3, then we can say
(-6)² -6 * b + c = 0
and
3² + 3b + c = 0
so:
36 - 6b + c = 0
and
9 + 3b + c = 0
Now just solve the two equations! Let's multiply the bottom one by 2:
18 + 6b + 2c = 0
then add it to the top one:
54 + 3c = 0
54 = -3c
-18 = c
Now plug c = -18 into 9 + 3b + c = 0 to solve for b:
9 + 3b - 18 = 0
3b = 9
b = 3
and we're done!
If our solutions are -6 and 3, then we can say
(-6)² -6 * b + c = 0
and
3² + 3b + c = 0
so:
36 - 6b + c = 0
and
9 + 3b + c = 0
Now just solve the two equations! Let's multiply the bottom one by 2:
18 + 6b + 2c = 0
then add it to the top one:
54 + 3c = 0
54 = -3c
-18 = c
Now plug c = -18 into 9 + 3b + c = 0 to solve for b:
9 + 3b - 18 = 0
3b = 9
b = 3
and we're done!
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Scott@TargetTestPrep
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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:BTGmoderatorLU wrote: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
The OA is B.
I'm confused with this PS question. Please, can any expert assist me with it? Thanks in advanced.
(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, and their sum is -15.
Answer: B
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