Solving $|a+1|+|2a+5|=5$

Question

How to solve this equation?

$$|a+1|+|2a+5|=5$$

Here is my solution but it seems that my answers are wrong.

$$|a+1|+|2a+5|=5$$

Case 1:

$$a+1+2a+5=5$$

$$a = \frac{1}{3}$$

Case 2:

$$a+1-2a-5=5$$

$$a=-11$$

Answer 1

HINT: There are three cases for solving $|a+1|+|2a+5|=5$, as follows

Case 1: $a< -\frac52$ $$-(a+1)-(2a+5)=5$$ Case 2: $-\frac52\le a\le -1$ $$-(a+1)+(2a+5)=5$$ Case 3: $a> -1$ $$(a+1)+(2a+5)=5$$ You can find feasible values of $a$ in above three cases.

Answer 2

$$|a+1|+|2a+5|=5````(1)$$ Three intervals of $a$ are to be explored. $R_1: d<-5/2$ Then (1) becomes $$-(a+1)+-(2a+5)=5 \implies a=-11/3.$$ $R_2:-5/2 \le a \le -1$ Then $$ -(a+1)+(2a+5)=5 \implies a=1 ~\text{which can not be found inthis interval)}$$ $R_3: a>-1$ Then $$(a+1)+2a+5=5 \implies a=-1/3$$ So only two solutions:$ -11/3, -1/3$.

Answer 3

Squaring both sides gives

$$(a+1)^2+|(a+1)(2a+5)|+(2a+5)^2=5^2,$$

or

$$|(a+1)(2a+5)|=25-(a^2+2a+1)-(4a^2+20a+25),$$

which gives

$$|(a+1)(2a+5)|=-5a^2-22a-1.$$

Squaring again, we obtain

$$(a+1)^2(2a+5)^2=(5a^2+22a+1)^2,$$

or

$$(2a^2+7a+5)^2=(5a^2+22a+1)^2,$$

or

$$(2a^2+7a+5)^2-(5a^2+22a+1)^2=0.$$

Factorising then gives

$$(-3a^2-15a+4)(7a^2+29a+6)=0,$$ from which you should be able to proceed.