Question 818956: The question asks to find a quadratic equation whose solutions are +2i and -2i. How do you find this with no other information?
Found 2 solutions by ewatrrr, TimothyLamb:
Answer by ewatrrr(24785)   (Show Source):
You can put this solution on YOUR website!
Hi
find a quadratic equation whose solutions are +2i and -2i
The factor theorem states that a polynomial f(x) has a factor (x − k) if and only if f(k) = 0.
f(x) = (x+2i)(x-2i) = x^2 + 4 |note i^2 = -1
Check answer with FOIL
F First terms x^2
O Outside terms -2i
I Inside terms +2i
L Last terms -4(-1)
Answer by TimothyLamb(4379)  (Show Source):
You can put this solution on YOUR website! ---
the standard form of a quadratic equation is:
f(x) = ax^2 + bx + c
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for this problem, the roots of the quadratic are given as a complex conjugate-pair:
+2i and -2i
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the fact that the roots are complex means the discriminant of the quadratic is negative.
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for details on the discriminant read the help text for "Discriminant" here:
https://sooeet.com/math/quadratic-formula-solver.php
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the roots may be calculated from the standard form given above, by using the quadratic formula.
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for details on the quadratic formula read the help text for "First quadratic root" here:
https://sooeet.com/math/quadratic-formula-solver.php
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the fact that the complex roots do not have a real part
(i.e. the complex roots only have an imaginary part)
tells us (from the quadratic formula), that b=0
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to see why this is so, read the help text given just above
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so now we can simplify the standard form to:
f(x) = ax^2 + (0)x + c
f(x) = ax^2 + c
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the discriminant of the quadratic is: b^2 - 4ac
but we now know that for this quadratic b=0
so the discriminant can be simplified to: -4ac
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so for this particular problem, all that remains of the quadratic formula is this:
x = (+)or(-) sqrt( -4ac )/2a
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and we know from the given roots that x= +2i and x= -2i
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so -4ac < 0 because that is where i comes from: i= sqrt(-1)
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given all of the above, solve for "a" in the simplified quadratic formula above:
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x = (+)or(-) sqrt( -4ac )/2a
x = (+)or(-) sqrt(-1) * sqrt(4ac)/2a
x = (+)or(-) i * sqrt(4ac)/2a
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use one of the given roots with the simplified quadratic formula above (either root will work, so x = +2i):
x = +2i = +i * sqrt(4ac)/2a
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sqrt(4ac)/2a = 2
sqrt(4ac) = 4a
4ac = 4a4a
c = 4a
a = c/4
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answer:
there is an infinite set of quadratic equations of the form f(x) = ax^2 + bx + c
where the roots of each equation are x= +2i and x= -2i
and where:
a = c/4
b = 0
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one such equation is this:
f(x) = x^2 + 4
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check:
the above quadratic equation is in standard form, with a=1, b=0, and c=4
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to solve the quadratic equation by using the quadratic formula, plug this:
1 0 4
into this: https://sooeet.com/math/quadratic-equation-solver.php
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this quadratic has complex roots:
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the two complex roots of the quadratic are:
x = 0 + (2)i
x = 0 - (2)i
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these are the same roots given in the problem statement, so our result is correct !
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