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?

  • 1
    $\begingroup$ 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 $-4+4.6=0.6$. $\endgroup$ – André Nicolas Sep 4 '12 at 14:16

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.

  • 1
    $\begingroup$ so the question says greater which means in my case it is 0.60 right? $\endgroup$ – JackyBoi Sep 4 '12 at 13:20
  • $\begingroup$ @JackyBoi You are correct. $\endgroup$ – Austin Mohr Sep 4 '12 at 13:21

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.

  • 2
    $\begingroup$ +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. $\endgroup$ – user22805 Sep 5 '12 at 3:15

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