If $f(x)=2|2x-5|+3$, with $x\geq0$, find all $k$ such that $f(x) = kx + 2$ has exactly two roots

Question

If $f(x) = 2|2x-5| + 3 $, $x\geq0$, find the values of $k$ such that the equation $f(x) = kx + 2$ has exactly two roots.

My attempt :-

$2(2x -5)+3 = kx + 2$ ,

$x \geq 2.5$

$x =\frac{9}{4-k}$

Then

$\frac{9}{4-k} \geq \frac{2}{5}$ , $k<4$

Then

$\frac{2}{5} < k < 4$

$2(-2x +5) + 3 = kx +2$ , $ x < \frac{2}{5}$ , $x\geq 0$

$x = \frac{11}{k+4}$

Then

$ \frac{11}{k+4} < \frac{5}{2}$

Then

$k > \frac{2}{5}$

Then

$\frac{2}{5} < k < 4$

Is there any easier procedure to find the values of $k$ ?

Answer 1

The set of lines $y = kx + 2$ are the non-vertical lines passing through the point $(0,2)$, while the graph of $y = 2|2x - 5| + 3 = 4|x - {5 \over 2}| + 3$ consists of two rays containing $({5 \over 2}, 3)$, one going upwards with slope $4$ and the other going upwards with slope $-4$.

In order for a line containing $(0,2)$ to intersect the graph of $y = 4|x - {5 \over 2}| + 3$ twice, it must (separately) intersect both of the above rays. Thus the minimal value of $k$ corresponding to this situation is when the line $y = kx + 2$ intersects the vertex $({5 \over 2}, 3)$, and the maximal value of $k$ is when the slope $k$ of the line is $4$, the slope of the right-hand ray. In both of these endpoint situations the the line will intersect the graph only once and thus shouldn't be included.

To find the minimum $k$, we simply plug $({5 \over 2}, 3)$ into $y = kx + 2$, obtaining $3 = {5 \over 2}k + 2$ or $k = {2 \over 5}$.

Thus the set of $k$ in question is given by ${2 \over 5} < k < 4$.

Answer 2

The equation is equivalent to: $$ 2\vert 2x−5\vert +3=kx+2 \land x\ge0\\ 2\vert 2x−5\vert=kx-1 \land x\ge0\\ (2\vert 2x−5\vert)^{2}=(kx-1)^{2} \land kx-1\ge0 \land x\ge0\\ (2 (2x−5))^{2}-(kx-1)^{2}=0 \land kx-1\ge0 \land x\ge0\\ (2(2x−5)+kx-1)(2(2x−5)-(kx-1))=0 \land kx-1\ge0 \land x\ge0\\ ((4+k)x−11)((4-k)x−9)=0 \land kx-1\ge0 \land x\ge0\\ (x=\frac{11}{4+k} \lor x=\frac{9}{4-k})\land kx-1\ge0 \land x\ge0\\ $$ So the equation have two solutions iff $$ \frac{11}{4+k}\neq\frac{9}{4-k}\land kx-1\ge0 \land x\ge0\\ 9(4+k)\neq\ 11(4-k)\land kx-1\ge0 \land x\ge0\\ k\neq\ \frac{2}{5}\land kx-1\ge0 \land x\ge0$$ now $kx-1\ge0$, but $(x=\frac{9}{4-k} \lor x=\frac{11}{4+k})$ so $$ \frac{9}{4-k}k-1\ge0 \land \frac{11}{4+k}k-1\ge0 \\ \frac{2}{5}\le \:k<4 \land (\:k<-4\lor k\ge \frac{2}{5}) $$ we also have that $x\ge0$, but $(x=\frac{9}{4-k} \lor x=\frac{11}{4+k})$ so $$ \frac{9}{4+k}\ge0 \land \frac{11}{4-k}\ge0\\ k\gt-4 \land 4\gt k $$ then $$ \frac{2}{5}\le \:k<4 \land (\:k<-4\lor k\ge \frac{2}{5})\land 4\gt k\gt-4 \land k\neq\ \frac{2}{5}\\ \frac{2}{5}\lt k<4 $$