2
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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
7
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.