1. Solve for modulus inequality.

Solve $\displaystyle \frac{|x^2-5x+4|}{|x^2-4|}\le1$

as

$\displaystyle |x^2-4|$will be positive always

cross multiply and take 1 to other side of equation
solve by taking LCM
we get
|x^2-5x+4|-(x^2-4)\le0
on solving we get

$\displaystyle (x^2-5x+4)-(x^2-4)\le0$ and $\displaystyle -(x^2-5x+4)-(x^2-4)\le0$

the other method I know is to square to remove the modulus function
$\displaystyle (x^2-5x+4)^2-(x^2-4)^2\le0$

among these which method is correct?
the second method becomes equation of degree 4 i.e.$\displaystyle x^4$)

2. Re: Solve for modulus inequality.

Originally Posted by sumedh
Solve $\displaystyle \frac{|x^2-5x+4|}{|x^2-4|}\le1$

as

$\displaystyle |x^2-4|$will be positive always

cross multiply and take 1 to other side of equation
solve by taking LCM
we get
|x^2-5x+4|-(x^2-4)\le0
on solving we get

$\displaystyle (x^2-5x+4)-(x^2-4)\le0$ and $\displaystyle -(x^2-5x+4)-(x^2-4)\le0$

the other method I know is to square to remove the modulus function
$\displaystyle (x^2-5x+4)^2-(x^2-4)^2\le0$

among these which method is correct?
the second method becomes equation of degree 4 i.e.$\displaystyle x^4$)
\displaystyle \displaystyle \begin{align*}\frac{|x^2 - 5x + 4|}{|x^2 - 4|} &\leq 1 \\ |x^2 - 5x + 4| &\leq |x^2 - 4| \\ (x^2 - 5x + 4)^2 &\leq (x^2 - 4)^2 \\ x^4-10x^3+33x^2-40x+16 &\leq x^4 - 8x^2 + 16 \\ 0 &\leq 10x^3 - 41x^2 + 40x \\ 0 &\leq x(10x^2 - 41x + 40) \\ 0 &\leq x\left(10x^2 - 16x - 25x + 40\right) \\ 0 &\leq x\left[2x(5x - 8) - 5(5x - 8)\right] \\ 0 &\leq x(5x - 8)(2x - 5) \end{align*}

Clearly when $\displaystyle \displaystyle f(x) = x(5x - 8)(2x - 5) = 0$ we have $\displaystyle \displaystyle x = 0$ or $\displaystyle \displaystyle x = \frac{8}{5}$ or $\displaystyle \displaystyle x = \frac{5}{2}$. Since the function changes sign at the x-intercepts, checking the function value at values of x in between the x-intercepts will tell us the intervals where the function is positive.

$\displaystyle \displaystyle f(-1) = -91$, so when $\displaystyle \displaystyle x < 0$ the function is negative.

$\displaystyle \displaystyle f\left(1 \right) = 9$, so when $\displaystyle \displaystyle 0 < x < \frac{8}{5}$, the function is positive.

$\displaystyle \displaystyle f(2) = -4$, so when $\displaystyle \displaystyle \frac{8}{5} < x < \frac{5}{2}$, the function is negative.

$\displaystyle \displaystyle f(3) = 21$, so when $\displaystyle \displaystyle x > \frac{5}{2}$, the function is positive.

So this inequality is satisfied when $\displaystyle \displaystyle x \in \left[0, \frac{8}{5}\right] \cup \left[\frac{5}{2}, \infty\right)$

3. Re: Solve for modulus inequality.

Thank you very much
I was making mistake in solving(X^2-4)
for checking the sign of the function
is it easy to put random values before, between and after the zeros to check the sign
or to make the sign table(attached)??

4. Re: Solve for modulus inequality.

Originally Posted by sumedh
Thank you very much
I was making mistake in solving(X^2-4)
for checking the sign of the function
is it easy to put random values before, between and after the zeros to check the sign
or to make the sign table(attached)??
I don't understand what you're doing with this "sign table". It is straightforward enough to substitute values that lie in the required interval to determine the nature of f(x).

5. Re: Solve for modulus inequality.

thank you very much