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I am trying to solve this. I already got the answer but my doubt is if I am doing it right.

There are two solutions for $x$ in the equation, to 2 decimal places. What is the value of the greater of the two solutions?

$a|x+b|+c = 0$ $a = 10$, $b = 4$ and $c = -46$

my solution's as follows:

Step 1, Substitute the values:

$10|x+4|-46 = 0$

Step 2, solve:

$10|x+4| = 46$

$|x+4| = +4.6$ or $|x+4| = -4.6$

Hence $x$ can be $x = 0.60$ or $x = -8.60$.

is that the correct way of doing it?

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    Note that $|x+4|=4.6$ can be rewritten as $|x-(-4)|=4.6$. Geometrically, this says that the **distance** of $x$ from $-4$ is $4.6$. The bigger solution is then $4.6$ to the right of $-4$, at $-4+4.6=0.6$.2012-09-04

2 Answers 2

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Your solutions are correct, but I have a quibble. It is not true that

$|x + 4| = 4.6 \text{ or } -4.6$

since the absolute value is always non-negative. Rather, you mean to say $ |x + 4| = 4.6 $ which implies $ x+4 = 4.6 \text{ or } x+4 = -4.6 $ (since applying the absolute value either did nothing or it removed the negative sign). Now you have two simpler equations to solve, which give $x = 0.6$ or $x = -8.6$ as solutions.

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    @JackyBoi You are correct.2012-09-04
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What you have done is correct. Note that when you have $|x+4|=4.6$, it means either $x \geq -4$ so your equation is $x+4=4.6$ or $x<-4$ so your equation will be $-(x+4)=4.6$ and you'll have to check if your answers meet the initial conditions which is the case in your question. I think that's what you had on your mind, just to make sure.

So $x=0.6$ is your final answer.

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    +1 for pointing out the importance of checking that the final solutions ARE in fact solutions to the initial equation. Failing to do this (for a whole range of different problems) would give an incomplete answer. So this answer is the better of those that have been posted so far.2012-09-05