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What is the relation between the coefficients of $ax^2+bx+c$ for both the roots to be equal and imaginary.

See we know that $b^2-4ac=0$ has roots equal and real.

This may be very elementary but its is bothering me for a few days.

Pole_Star
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  • Can you find an example that the above is true indeed? Because at me seems that the only case is the one you've already mentioned. –  Apr 21 '17 at 11:50
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    You can still have $b^2-4ac=0$ give a double complex root if either $a$ or $b$ is a complex number! (Or potentially both in most cases...) :) – Cameron Williams Apr 21 '17 at 11:51

4 Answers4

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We denote the roots by $x_1$ and $x_2$. If they are equal and imaginary, then there is a real $t \ne 0$ such that $x_1=x_2=it$. Hence

$ax^2+bx+c=a(x-it)^2=ax^2-2itax-at^2$.

Your turn !

Fred
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The discriminant needs to be $0$. Thus, we need to solve $b^2=4ac$, but $a,b,c$ can be complex rather than real. For example, we could do $b=-2i$, $a=1$ and $c=-1$. We see

$$x^2-2ix-1=0$$

has solution $i$ twice (since $(x-i)^2=x^2-2ix-1$).

Note though that it is easy to construct a polynomial from its roots, so if we wish to have a quadratic with two equal roots, then $\beta(x-\alpha)^2=\beta x^2-2\alpha\beta x+\alpha^2\beta=0$ suffices, and one can simply choose $\alpha$ and $\beta$ complex.

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Hint:$$a{ \left( x-i \right) }^{ 2 }=0\\ a{ x }^{ 2 }+2aix-a=0\\ b=2ai,c=-a$$

haqnatural
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The rule "$b^2-4ac=0$ has roots equal and real" comes from the study of real polynomials, where $a,b,c$ all are real numbers. In that case you cannot have equal imaginary roots.

For complex polynomials, it is not always true that the quadratic $ax^2 + bx + c,$ where $b^2-4ac=0,$ has roots that are equal and real. If $b^2-4ac=0$ then the roots are equal, but they could be equal to any complex number.

So you need not only a rule for telling when the roots are equal and imaginary, you must also have a new rule for telling when the roots are equal and real. Consider "$b^2-4ac=0$ and $a,b,c$ all real," for example. That is a sufficient but not necessary set of conditions to have two real roots; when it is true, the roots will be equal and real (no false positives), but there are many equations that have equal real roots even though $a,b,c$ are not all real (lots of false negatives). We should be able to come up with a better rule than that to tell when a complex quadratic has equal real roots.

Each of these rules will include the condition $b^2-4ac=0$ plus some other conditions. To find the other conditions, examine the other answers to your question. You might consider putting a condition on the value of $\frac ba$ and another on the value of $\frac ca.$

David K
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