For what values of $a,b$ does the equation $${ x }^{ 2 }+2\left( 1+a \right) x+\left( 3{ a }^{ 2 }+4ab+4{ b }^{ 2 }+2 \right) = 0$$ have real roots?

For it to have real roots, the discriminant has to be $>0$, correct? (Or equal to, I suppose, since the question didn't specify distinct or not) So I tried using the values, which gave me ${ \left( 2+2a \right) }^{ 2 }-4\left( 3{ a }^{ 2 }+4ab+4{ b }^{ 2 } +2\right) $ but I'm not sure where to go after that.

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    $\begingroup$ You are correct that the discriminant has to be greater than zero, in order to have two real roots (assuming $a,b$ are real coefficients). You would get a double real root if the discriminant was exactly zero. $\endgroup$ – hardmath Aug 7 '15 at 21:20

You have the right idea. You just need to continue expanding that expression.

From $x^2 + 2(1+a)x + (3a^2 + 4ab + 4b^2 + 2) = 0 $, the discriminant is (ignoring the factor of 2 since we are concerned only about the sign)

$\begin{array}\\ d &=(1+a)^2-(3a^2 + 4ab + 4b^2 + 2)\\ &=a^2+2a+1-(3a^2 + 4ab + 4b^2 + 2)\\ &=-2a^2+2a-1-4ab-4b^2\\ &=-a^2+2a-1-a^2-4ab-4b^2 \quad\text{This is the key step}\\ &=-(a-1)^2-(2b+a)^2\\ \end{array} $

So, in the miraculous way of many homework problems, this is the negative of a sum of two squares.

So $d \le 0$ and, for $d = 0$, we must have $a=1$ and $2b+a=0$, which means $b = -1/2$.

For any other values of $a$ and $b$, the discriminant is negative, and so there are no real roots.

For these values of $a$ and $b$, there is a repeated root.

  • $\begingroup$ In general I don´t agree, that ignoring the factor 2 is right here. Could you give me more explanation why it is allowed here. $\endgroup$ – callculus Aug 7 '15 at 21:27
  • $\begingroup$ Could you explain why we can ignore the factor of 2? Also, after that if it's correct, that means there are NO values of a,b in which the equation has 2 real roots right? Well, the question didn't specify distinct though, so the answer must be the repeated root. $\endgroup$ – mathflair Aug 7 '15 at 21:28
  • $\begingroup$ You are searching for when $\Delta\ge0$, but you have $\Delta\ge0\iff k\Delta\ge0$ where $k$ is positive $\endgroup$ – Elliot G Aug 7 '15 at 21:30
  • $\begingroup$ $b^2-4ac = 4((b/2)^2 - ac)$, so the sign of $b^2-4ac$, which is what we are interested in, is the same as the sign of $(b/2)^2-ac$. $\endgroup$ – marty cohen Aug 7 '15 at 21:31
  • $\begingroup$ Sorry Elliot, can you elaborate? $\endgroup$ – mathflair Aug 7 '15 at 21:35

Here it is easy to complete the square $$(x+1+a)^2+3a^2+4ab+4b^2+2-(1+a)^2=0$$ so that $$(x+1+a)^2=-2a^2+2a-1-4b^2-4ab=-(a-1)^2-(2b+a)^2$$

This step involves simply completing the square for $4b^2+4ab=(2b+a)^2-a^2$ and seeing what is left.

The left-hand side is non-negative, the right-hand side non-positive, so equality is only possible if both are zero.

Note: this is wholly equivalent to working with the discriminant, but saves a factor of $4$ and does not require remembering a formula.


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