SOLUTION: i'm trying to help my daughter with homework over the phone
she is trying to solve the equation x^2 + bx + c = 0
when x = -4 or 6, what are b and c
haven't a clue where to start
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-> SOLUTION: i'm trying to help my daughter with homework over the phone
she is trying to solve the equation x^2 + bx + c = 0
when x = -4 or 6, what are b and c
haven't a clue where to start
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Question 392768: i'm trying to help my daughter with homework over the phone
she is trying to solve the equation x^2 + bx + c = 0
when x = -4 or 6, what are b and c
haven't a clue where to start Found 4 solutions by Alan3354, stanbon, ewatrrr, solver91311:Answer by Alan3354(69443) (Show Source):
You can put this solution on YOUR website! x^2 + bx + c = 0
when x = -4 or 6, what are b and c
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x = -4 or 6 --> (x + 4)*(x - 6) = 0
-->
b = -2
c = -24
You can put this solution on YOUR website! solve the equation x^2 + bx + c = 0
when x = -4 or 6, what are b and c
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If x = -4 you get: (-4)^2+b(-4)+c = 0
16-4b+c = 0
-4b+c = -16
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If x = 6, you get: (6)^2+b(6)+c = 0
6b+c = -36
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Now you have 2 equations with variables b and c.
-4b+c = -16
6b+c = -36
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Subtract the top equation from the lower one and solve for "b":
10b = -20
b = -2
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Solve for c by substituting into 6b+c = -36:
6*-2+c = -36
-12+c = -36
c = -24
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Check in the equation -4b+c = -16
-4*-2+(-24) = -16
8-24 = -16
-16 = -16
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Cheers,
Stan H.
Hi
Given x = -4 or 6, those are the 'roots' of the quadratic equation and
Note:
y = (x - (-4))(x-6)
y = (x+4)(x-6) |multiplying gives us
y = x^2 + 4x - 6x - 24
y = x^2 - 2x - 24 = 0
therefore, in this example of a standard quadratic equation y = x^2 + bx + c
b = -2 and c = -24
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