If the square of the sum of the roots of a quadratic equation

For simplicity I'll assume the quadratic is in the form x^2 + kx + m = 0, but it's easy to adapt the below to the case where there is a number in front of the "x^2"; it's not worth the trouble since this question is nothing like a real GMAT question.

If a and b are the roots (solutions) of that quadratic, then (x - a)(x - b) = 0. The question tells us (a + b)^2 < ab. The left side of this inequality is clearly positive, so if the right side ab is larger than the left side, then ab must also be positive. But now if we expand, we get

a^2 + b^2 + 2ab < ab
a^2 + b^2 < -ab

and a^2 + b^2 is positive, since we're adding two squares, but -ab is negative, since ab is positive. A positive number can't be less than a negative number, so this is impossible.

So if we're working with real numbers, there's no way this could happen. It could only happen if non-real ('imaginary' or 'complex') numbers are involved. Imaginary numbers are absolutely never tested on the GMAT -- the test instructions state that explicitly -- so this is not a remotely realistic question.

(-b/a)^2 < c/ a => b^2 < ac ------ 1

as we know for imaginary numbers : b^2 - 4ac < 0 thus b^2< 4ac ----------- 2
if it is less than ----- 1
definitely less than ----- 2


Hope this helps!!

The sum of the roots of a quadratic equation is \(X+Y=\frac{-b}{a}\) and the product is \(XY=\frac{c}{a}\), where \(a\) and \(b\) are the coefficients of \(x^2\) and \(x\) respectively and c is the constant term.

The question states \((X+Y)^2 < XY\)
\((\frac{-b}{a})^2 < \frac{c}{a}\)

Now we can multiply \(a\) on both sides. The problem is we don't know if \(a\) is positive or negative. Lets examine both cases.

If \(a > 0\) :
\(\frac{(-b)^2}{a} < c\)
Multiply both sides by \(a\) again.
\((-b)^2 < ac\)

If \(a < 0\) :
\(\frac{(-b)^2}{a} > c\)
Multiply both sides by \(a\) again.
\((-b)^2 < ac\)

Both cases give us the same information.

One of the options says the roots are not real. It makes sense to check that first by finding the discriminant (\(D\)), because, we have information relating to \(b^2\) and \(ac\), both of which show up in the formula for the discriminant.

\(D = b^2 - 4ac\)

We now know that \((-b)^2\) (which is the same as \(b^2\)) is less than \(ac\). So \(b^2\) obviously will be less than \(4ac\).

Therefore \(D < 0\).

Roots are not real.