# Modulus (absolute value) function

Find all values of a where 'a' belongs to all real numbers, for the equation a³ + a²|a + x| +|a²x + 1| = 1 has not less than four different solutions which are integers. X is a variable Please explain everything

• I'm unclear on your question. Do you want to find integer values of "x" for which your equation is true for all real values of "a"? Also, please explain why you think there are at least four different solutions. Thanks. – John Omielan Jan 25 at 12:13
• Its the question in my textbook – tapus aggarwal Jan 25 at 12:20

$$1-a^3\geq0,$$ which gives $$a\leq1.$$

We see that $$a=1$$ and $$a=0$$ are valid.

Let $$a\neq0$$ and $$a\neq1$$.

Thus, $$-\frac{1}{a^2}<-a$$ and we have three cases:

1. $$x>-a$$.

We have here $$a^3+a^2x+a^3+a^2x+1=1$$ or $$x=-a,$$ which is not good.

1. $$-\frac{1}{a^2}\leq x\leq-a.$$

We obtain $$a^3-a^2x-a^3+a^2x+1=1,$$ which is very well.

1. $$x<-\frac{1}{a^2}.$$

Can you end it now?

• No that's not what is asked for, values of a are asked or the range of a not x – tapus aggarwal Jan 25 at 15:46
• @tapus aggarwal I know. The answer is $a\leq1$. It follows from my solution. – Michael Rozenberg Jan 25 at 16:29