Question 696524: in a quadratic equation x^2 + px + q =0 if sum of its roots are equal to thrice of their difference then proove that
2p^2 = 9q Found 2 solutions by stanbon, Edwin McCravy:Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! in a quadratic equation x^2 + px + q =0 if sum of its roots are equal to thrice of their difference then prove that 2p^2 = 9q
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Roots:
[-p+sqrt(p^2-4q)]/(2)
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[-p-sqrt(p^2-4q)]/(2)
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Sum: -p/2+-p/2 = -p
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Diff: sqrt(p^2-4q)
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Equation:
sum = 3*diff
-p = 3[sqrt(p^2-4q)]
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sqrt(p^2-4q) = -p/3
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Square both sides to get:
p^2 - 4q = p^2/9
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9p^2 - 36q = p^2
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8p^2 = 36q
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2p^2 = 9q
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Cheers,
Stan H.
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In this approach we don't actually find the roots as the other
tutor did, although that method is quite correct.
x² + px + q = 0
Let the roots be r1 and r2.
Then the above factors as
(x - r1)(x - r2) = 0
or, upon "FOIL"ing that out:
x² - r2x - r1x + r1r2 = 0
x² - (r2+r1)x + r1r2 = 0
x² - (r1+r2)x + r1r2 = 0
comparing this to
x² + px + q = 0
p = -(r1+r2)
q = r1r2
Now we use the fact that the sum of its roots
are equal to thrice their difference, which
says:
r1+r2 = 3(r1-r2) or r1+r2 = 3(r2-r1)
Both equations are technically necessary because we are not told
which order we are to subtract the roots to find the difference.
r1+r2 = 3r1-3r2 or r1+r2 = 3r2-3r1
4r2 = 2r1 or 4r1 = 2r2
2r2 = r1 or 2r1 = r2
But since they are the same except for which root we call
r1 and which we call r2, we can just take the first case.
Now we have these 3 equations:
p = -(r1+r2)
q = r1r2
2r2 = r1
Using the third equation, we substitute 2r2 for r1 in the
first two equations:
p = -(2r2+r2)
q = 2r2r2
Simplifying,
p = -3r2
q = 2r2²
Solve the first one for r2:
r2 =
r2 =
Substitute for r2 in q = 2r2²:
q = 2
q = 2
q = 2
Multiply both sides by 9
9q = 2p²
That's the same as
2p² = 9q
Edwin
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