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!
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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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Solve and graph linear equations:

https://sooeet.com/math/linear-equation-solver.php

---

Solve quadratic equations, quadratic formula:

https://sooeet.com/math/quadratic-formula-solver.php

---

Calculate and graph the linear regression of any data set:

https://sooeet.com/math/linear-regression.php 

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