if $mx+3|x+4|-2=0$ has no solutions, solve for $m$

Question

If $mx+3|x+4|-2=0$ has no solutions, which of the following value could be $m$?

(A)5

(B)$-\frac{1}{2}$

(C)-3

(D)-6

(E)$\frac{10}{3}$

my attempt: $$mx-2=-3|x+4| \\ m^2x^2-4mx+4=9x^2+72x+144 \\ (9-m^2)x^2+(4m+72)x+140=0$$ because the equation has no solutions, therefore $$(4m+72)^2-4(9-m^2)140<0 \\ 576m^2+576m+144<0 \\ 4m^2+4m+1<0 \\ (2m+1)^2<0$$ maybe I made a mistake but I couldn't find it

Answer 1

The hint.

For $x\geq-4$ we obtain: $$mx+3x+12-2=0,$$ which gives a value $m=-3$.

For $x\leq-4$ we obtain $$mx-3x-12-2=0,$$ which gives a value $m=3.$

Now, check that for $m=3$ our equation has root, while for $m=-3$ our equation has no roots.

Answer 2

1) $y_1=3|x-(-4)| \ge 0$ ;

2) $y_2= Mx +2$, where $M=-m$

$M=3=-m$ no intersection (Why?)

No intersection for $3 \ge M >1/2$ (Why?);

$3 \ge -m >1/2$, or

$-3 \le m < -1/2$.

Answer 3

Consider the fact that $|x+4|$ has 2 different possible values. For $x\geq-4$, it becomes $x+4$. For $x<-4$, it becomes $-x-4$. Divide the problem into 2 those two different cases.

For $x\geq-4$, the equation becomes $mx+3(x+4)-2=mx+3x+10=(m+3)x+10=0$. We obtain that $x=-\frac{10}{m+3}$. In order to make there exist no solution for $x$, the value of denominator shall be zero. Thus $m+3=0$ which implies that $m=-3$.

For $x<-4$, the equation becomes $mx+3(-x-4)-2=mx-3x-14=(m-3)x-14=0$. We obtain that $x=-\frac{14}{m-3}$. Similarly, in order to make there exist no solution for $x$, the value of denominator shall be zero. Thus $m-3=0$ which implies that $m=3$.

After getting two different values of $m$ from both cases, we shall check whether it is correct or not by inserting each value of $m$ and try to solve the equation. After checking the value of $m$, the only value that fulfill the requirement is $m=-3$.