For what values of $ a, b$ does the equation have real roots?

Question

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.

Answer 1

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.

Answer 2

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.