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• Sep 30th 2013, 04:19 PM
sakonpure6
Hello Can some one please check if i did this question right? Thank you.

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

so, -(x-4)^2 = 0 and (x-1)^2 =0

so, x=4 and x =1

there fore 1<x<4

Is this correct?
• Sep 30th 2013, 04:46 PM
Plato
Quote:

Originally Posted by sakonpure6
Hello Can some one please check if i did this question right? Thank you.
$\displaystyle -(x-4)^2 (x-1)^2 < 0$
so, -(x-4)^2 = 0 and (x-1)^2 =0
so, x=4 and x =1
there fore 1<x<4
Is this correct?

No it is far from correct. $\displaystyle -(x-4)^2 (x-1)^2$ is always non-positive.
The solution set is $\displaystyle \mathbb{R}\setminus\{1,4\}$, all real numbers except 1 & 4.
• Sep 30th 2013, 08:39 PM
Prove It
Quote:

Originally Posted by Plato
No it is far from correct. $\displaystyle -(x-4)^2 (x-1)^2$ is always non-positive.
The solution set is $\displaystyle \mathbb{R}\setminus\{1,4\}$, all real numbers except 1 & 4.

The reason being is because x = 1 and x = 4 are the only values for which the expression actually equals 0, which is NOT less than 0.
• Oct 1st 2013, 05:13 AM
Hartlw
It's a quartic inequality, not quadratic. Sure it's not -(x-4)(x-1)<0?
• Oct 1st 2013, 06:04 AM
Plato
Quote:

Originally Posted by Hartlw
It's a quartic inequality, not quadratic. Sure it's not -(x-4)(x-1)<0?

A rose by any other name is still a rose.
• Oct 1st 2013, 06:20 AM
Hartlw
whoops. I missed the point. It is a quadratic:

(x-4)^2(x-1)^2>0, so now you can take the square root to get requirement (x-4)(x-1)>0.
• Oct 1st 2013, 07:28 AM
thevinh
Quote:

Originally Posted by Hartlw
whoops. I missed the point. It is a quadratic:

(x-4)^2(x-1)^2>0, so now you can take the square root to get requirement (x-4)(x-1)>0.

Let's take a step back and think about the problem. $\displaystyle (x-4)^2(x-1)^2>0$ is always greater than zero since everything is being squared. So the answer to this problem is ALL REAL VALUES except $\displaystyle x=4$ and $\displaystyle x=1$.

Now if the question is $\displaystyle (x-4)^2(x-1)^2 \geq 0$, then the answer will be ALL REAL VALUES ONLY. Can you see the point here?
• Oct 1st 2013, 07:39 AM
Hartlw
Quote:

Originally Posted by thevinh
Let's take a step back and think about the problem. $\displaystyle (x-4)^2(x-1)^2>0$ is always greater than zero since everything is being squared. So the answer to this problem is ALL REAL VALUES except $\displaystyle x=4$ and $\displaystyle x=1$.

Now if the question is $\displaystyle (x-4)^2(x-1)^2 \geq 0$, then the answer will be ALL REAL VALUES ONLY. Can you see the point here?

I don't see your point. The problem starts with: -(x-4)^2 (x-1)^2 < 0
It's pretty obvious there's a catch because when you first look at it you see a minus sign and you don't think of taking the square root. They're telling you to multiply by -1 to make it positive and then you can take the square root, which is positive by convention, and that gives the quadratic inequality which is the topic of the thread.

As is, it is trivial.
• Oct 1st 2013, 10:02 PM
thevinh
Quote:

Originally Posted by Hartlw
I don't see your point. The problem starts with: -(x-4)^2 (x-1)^2 < 0
It's pretty obvious there's a catch because when you first look at it you see a minus sign and you don't think of taking the square root. They're telling you to multiply by -1 to make it positive and then you can take the square root, which is positive by convention, and that gives the quadratic inequality which is the topic of the thread.

As is, it is trivial.

Like you said, first multiple both side by -1 to obtain $\displaystyle (x-4)^2 (x-1)^2 > 0$. The point here is WE DON'T HAVE TO TAKE SQUARE ROOT, because it is obvious that anything square will always be greater than zero, except for zero itself. So the answer is ALL REAL VALUE EXCEPT 4 & 1. Beside taking a square on both side of the inequality is a dangerous maneuver. For example, would you take the square root on both side when solving this inequality: $\displaystyle (x+2)^2<1$?

Why do extra work to get the answer? Just simply think about the problem, analyze it, extract the most out of it.
• Oct 1st 2013, 10:08 PM
Prove It
Quote:

Originally Posted by thevinh
Like you said, first multiple both side by -1 to obtain $\displaystyle (x-4)^2 (x-1)^2 > 0$. The point here is WE DON'T HAVE TO TAKE SQUARE ROOT, because it is obvious that anything square will always be greater than zero. So the answer is ALL REAL VALUE.

No it's not. When x = 4 or x = 1, the expression = 0, which is NOT less than 0.

The solution is \displaystyle \displaystyle \begin{align*} x \in \mathbf{R} \backslash \left\{ 1, 4 \right\} \end{align*}.
• Oct 1st 2013, 10:38 PM
thevinh
Quote:

Originally Posted by Prove It
No it's not. When x = 4 or x = 1, the expression = 0, which is NOT less than 0.

The solution is \displaystyle \displaystyle \begin{align*} x \in \mathbf{R} \backslash \left\{ 1, 4 \right\} \end{align*}.

Sorry, I meant ALL REAL VALUE EXCEPT {1,4}. Thanks.
• Oct 3rd 2013, 04:40 AM
Hartlw
thevinh. You didn't finish the inequality. Assume:

0<(x+2)^2<1
0< x+2 <1
-2 < x < -1
• Oct 3rd 2013, 06:22 AM
Hartlw
a^2>0 is either trivial (a='0), or a>0. Topic of thread tells you which.

Look at it another way:
Let 0<a^2<N, then 0<a<sqrtN.
Let N -> infinity, then 0<a.
• Oct 3rd 2013, 06:37 AM
thevinh
Quote:

Originally Posted by Hartlw
thevinh. You didn't finish the inequality. Assume:

0<(x+2)^2<1
0< x+2 <1
-2 < x < -1

Hartlw. For the example $\displaystyle (x+2)^2<1$, would it still be true if x=-2.5? The right answer is $\displaystyle -3<x<-1$. The point is that it is a fatal mistake to take square root both sides when solving an inequality. The best way to solve this problem is to expand them, bring everything to one side, factor and solve. See below.

$\displaystyle (x+2)^2<1~ \Leftrightarrow~ x^2+4x+4<1~ \Leftrightarrow~ x^2+4x+3<0~ \Leftrightarrow~ (x+1)(x+3)<0$

This is a proper way of solving inequality that ensures no lost of solution.

From your solution, when x=-2 the inequality still holds.
• Oct 3rd 2013, 06:56 AM
Plato
Quote:

Originally Posted by thevinh
Hartlw. For the example $\displaystyle (x+2)^2<1$, would it still be true if x=-2.5? The right answer is $\displaystyle -3<x<-1$. The point is that it is a fatal mistake to take square root both sides when solving an inequality. The best way to solve this problem is to expand them, bring everything to one side, factor and solve. See below.
$\displaystyle (x+2)^2<1~ \Leftrightarrow~ x^2+4x+4<1~ \Leftrightarrow~ x^2+4x+3<0~ \Leftrightarrow~ (x+1)(x+3)<0$
This is a proper way of solving inequality that ensures no lost of solution.[/B]

I disagree with there being one proper way of solving anything.

For example: $\displaystyle {a^2} \le {b^2} \leftrightarrow |a| \le |b|$.

So in this case $\displaystyle (x+2)^2<1 \leftrightarrow |x+2|< 1$ or $\displaystyle -3<x<-1$.
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