Absolute value (1 Viewer)

Green Yoda · Mar 19, 2016

How to do this question?
x-3/|x-2|=4

InteGrand · Mar 19, 2016

Rathin said:
How to do this question?
x-3/|x-2|=4

$\noindent Note that $x\neq 2$. Now, if $x>2$, we have $\frac{x-3}{x-2}=4,x>2$. Equivalently, $x-3 = 4x - 8 \Longleftrightarrow 3x = 5 \Longleftrightarrow x = \frac{5}{3}$. But this is not greater than $2$. So we have no solutions for $x>2$.$

$\noindent Now if $x<2$, we have $\frac{x-3}{2-x}=4,x<2$. Equivalently, $x-3 = 8-4x \Longleftrightarrow 4x = 11 \Longleftrightarrow x = \frac{11}{4}$. But this is not less than 2. So there are no solutions for $x<2$. So overall, there are no real solutions.$

Green Yoda · Mar 19, 2016

InteGrand said:
$\noindent Note that $x\neq 2$. Now, if $x>2$, we have $\frac{x-3}{x-2}=4,x>2$. Equivalently, $x-3 = 4x - 8 \Longleftrightarrow 3x = 5 \Longleftrightarrow x = \frac{5}{3}$. But this is not greater than $2$. So we have no solutions for $x>2$.$

$\noindent Now if $x<2$, we have $\frac{x-3}{2-x}=4,x<2$. Equivalently, $x-3 = 8-4x \Longleftrightarrow 4x = 11 \Longleftrightarrow x = \frac{11}{4}$. But this is not less than 2. So there are no solutions for $x<2$. So overall, there are no real solutions.$

Yup I got the answer, but for the positive case why does x>2 and for the negative case why does it have to be x<2?

kawaiipotato · Mar 19, 2016

Rathin said:
Yup I got the answer, but for the positive case why does x>2 and for the negative case why does it have to be x<2?

Because for |x-2|, for any number x>2 then x-2 > 0
and since x-2 > 0 then |x-2| = +(x-2)
If x<2 then x-2<0 so |x-2| = -(x-2) = 2-x

braintic · Mar 19, 2016

Alternatively ..... divide the interval from 3 to 2 internally and externally in the ratio 4:1.
Since neither of these points lie to the right of 3, no solution.

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Absolute value (1 Viewer)

Green Yoda

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InteGrand

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Green Yoda

Hi Φ

kawaiipotato

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braintic

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