Currently, I am having trouble with the following questions listed below:
Solve the equation $$\left\lvert x-2\right\rvert -\left\lvert x+ 3\right\rvert =x^2 - 1$$
For this question, I have drawn the graph for both, and found that there are two points of intersection and hence two solutions.
I found that the point $-1$ was a point of intersection, but when I tried subbing it back into the equation, the answers were not equal, leaving me confused. I was unable to find the other answer, although I am aware that it is less than -1(due to graphing it)
Consider the equation $$\left\lvert x^2+3x+2 \right\rvert = 1 $$
Find the values for $x$ when $$ x^2+3x+2 >0$$ and those for which $$ x^2+3x+2 <0$$
For this part, I first factorized $x^2+3x+2$. I got $(x+2)(x+1)$.
For the greater than zero, I got negative $1$ and for less than I got $-2$.Hence, using the definition of absolute value, solve $\left\lvert x^2+3x+2 \right\rvert=1$.
I am somewhat confused by this part. I am not sure how to use the definition of a abolute value to solve the above, so I tried using the quadratic formula instead, which gave me the answer: $x=\dfrac{-3+\sqrt{5}}2$ or $x=\dfrac{-3-\sqrt{5}}2$
Thank you for your time