1. ## Inequality problem

The question is , "Solve the inequality 2 - 3x < |x -3|" .

i tried it this way.
squaring both sides, ( 2 - 3x)² < (x - 3)²

8x² - 6x - 5 < 0

equating it to zero, 8x² - 6x - 5 = 0

critical values are, x = -0.5 x= 5/4

using these values, i did a sketch of the curve , and as i was required to find values of x which are < 0 , since 8x² - 6x - 5 < 0 , i got the range as -0.5 < x < 5/4
But the correct answer is x > = - 0.5 only. Why is 5/4 not accepted ?

2. ## Re: Inequality problem

5/4 is accepted as it is > -0.5

3. ## Re: Inequality problem

Your formating has me a wee bit confused... Is the top supposed to be an absolute value? If so that has to carry down, you can't change it to parentheses later. Is (") supposed to be at the end of the absolute value?

4. ## Re: Inequality problem

the answer should be -0.5 <= x <= 5/4

as past 5/4 on real line would make inequality > 0

5. ## Re: Inequality problem

Originally Posted by nks2427
The question is , "Solve the inequality 2 - 3x < |x -3|" .
Originally Posted by Nappy
the answer should be -0.5 <= x <= 5/4
as past 5/4 on real line would make inequality > 0
The answer is $(-0.5,\infty).$

6. ## Re: Inequality problem

hey alane1994 , what do you mean when saying "the top" ? I just put the question in between quotes. There are no (") at the end of the absolute value as such.
i dont understand why 5/4 is not accepted.
In the answer, it says, "Fully Justify x > -0.5 as the only answer" :?

7. ## Re: Inequality problem

Originally Posted by nks2427
The question is , "Solve the inequality 2 - 3x < |x -3|" .
I will show you how to do this. Consider two cases.

If $x<3$ then solve $2-3x<3-x$ and get $-0.5.

If $x\ge 3$ then solve $2-3x< x-3$ and get $\frac{5}{4}< x$.

Union those two and get $(-0.5,\infty)$

8. ## Re: Inequality problem

okayy... but i need to give a range...so i should just put x > -0.5 then?

9. ## Re: Inequality problem

I feel like such an idiot... I didn't even notice that it was in quotes. I just thought it was a random quote. By top I meant first line. Ad for the range he gave it. I guess you haven't learned interval notation?