# Thread: Different methods of solving Inequalities

1. ## Different methods of solving Inequalities

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?

2. ## Re: Different methods of solving Inequalities

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

$(4x^2-1)(x^2-4) < 0$

critical values at $x = \pm \dfrac{1}{2}$ and $x=\pm 2$

$x \in (-\infty,-2) \implies y > 0$

$x \in \left(-2, -\dfrac{1}{2}\right) \implies y < 0$

$x \in \left(-\dfrac{1}{2}, \dfrac{1}{2} \right) \implies y > 0$

$x \in \left(\dfrac{1}{2}, 2 \right) \implies y < 0$

$x \in (2,\infty) \implies y > 0$

$x^4-\dfrac{17}{4}x^2+ 1 < 0$

$x^4-\dfrac{17}{4}x^2+ \dfrac{289}{64} - \dfrac{289}{64}+1 < 0$

$\left(x^2-\dfrac{17}{8}\right)^2 -\dfrac{225}{64} < 0$

$\bigg[\left(x^2-\dfrac{17}{8}\right)-\dfrac{15}{8}\bigg] \bigg[\left(x^2-\dfrac{17}{8}\right)+\dfrac{15}{8}\bigg] < 0$

$(x^2-4)\left(x^2-\dfrac{1}{4}\right) < 0$ ... which leads to the same intervals above

3. ## Re: Different methods of solving Inequalities

yes, I see the mistake.

I said since $x^2 < b^2$ then $x < b$ only considering the positive root.

thanks.

4. ## Re: Different methods of solving Inequalities

Originally Posted by s_ingram
yes, I see the mistake.

I said since $x^2 < b^2$ then $x < b$ only considering the positive root.

thanks.
Yes it should be |x| < |b|.