Hi folks,

When asked to find the range of values x can take given an inequality, there are at least three methods:

Sketching the function, completing the square or checking the sign of the factors. All three should agree, naturally.

So, given $y = 4x^4 - 17x^2 + 4 < 0 $

let $y = x^2$ then

$4y^2 - 17y + 4 < 0$

$y^2 -\frac{17}{4}y + 1 < 0$ ...................................(1)

$(4y - 1)(y - 4) < 0 $

so, examining the factors:

$

\begin{array}{lccc}

& y < \frac{1}{4} & \frac{1}{4} < y < 4 & y > 4 \\

4y - 1 & - & + & + \\

y - 4 & - & - & + \\

f(y) & + & - & +

\end{array}

$

we get $ \frac{1}{4} < y < 4$

resubstituting, $\frac{1}{4} < x^2 < 4$

and a little simplification yields the range:

$-2 < x < -\frac{1}{2} , \frac{1}{2} < x < 2$

Now,

if I decided to complete the square:

$y^2 -\frac{17}{4}y + 1 < 0$ then from (1)

we get

$(y - \frac{17}{8})^2 -(\frac{17}{8})^2 + 1 < 0$

i.e.

$(y -\frac{17}{8})^2 < \frac{289}{64} -\frac{64}{64} $

which gives

$(y - \frac{17}{8}) < \frac{15}{8}$

whence

$y < 4$ or $x^2 < 4$

giving the range

$ -2 < x < 2$ which does not match the result above.

clearly there is an arithmetic error somewhere. I can't see it, can anyone else?