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A student writes the below argument:

Consider
$a x^{4} + b x^{2} + c = 0$
If
$b^{2} - 4 a c > 0$
then the equation has four, distinct, real solutions.

Find a counterexample to the student's claim.

Hint

Whilst it is true that

\begin{aligned} x^{2} = α & \Rightarrow x = \pm \sqrt{α} \\ x^{2} = β & \Rightarrow x = \pm \sqrt{β} \end{aligned}

might provide four solutions, what if $α < 0$ ?

Solution

Consider, for example, $(u + 1) (u + 2) = u^{2} + 3 u + 2$ . With $u = x^{2}$ this suggests the equation

x^{4} + 3 x^{2} + 2 = 0

$b^{2} - 4 a c = 1 > 0$ but the solutions would be given by

\begin{aligned} x^{2} & = - 1 \\ x^{2} & = - 2 \end{aligned}

neither of which are possible for $x \in R$ .

So $x^{4} + 3 x^{2} + 2 = 0$ has no real solutions!

TIP

There are infinitely many possible counterexamples - you just need to give one and it does not need to be the one in the solution.