Let $k$ be a real number. Prove that if the equation $|x^{2} - 3x| = x-2+k$ has two distinct roots, then either $-1 < k < 2$ or $k > 3$?

Question

The title is the problem.

The condition "has two distinct roots" is ambiguous, but I assume it to be ``having exactly two distinct roots".

This is false. Take $k=3$; then it has roots at $1,2\pm \sqrt{5}$. — pre-kidney, Feb 08 '15 at 07:25

score 2 · Accepted Answer · answered Feb 08 '15 at 07:39

2

Here are the number of real roots as $k$ varies: $$ \begin{cases} 0,&k<-1\\ 1,&k=-1\\ 2,& -1<k<2\\ 3,& k=2\\ 4,& 2<k<3\\ 3,& k=3\\ 2,& k>3 \end{cases} $$

To see why, plot $|x^2-3x|-x+2$:

Plot of |x^2-3x|-x+2

Then the number of real roots for a given $k$ corresponds to the number of points in the horizontal slice at height $k$ of this graph.

For a formal proof, you can split up into cases based on if $x\in [0,3]$ or not; then the equation splits up into two quadratic equations.

answered Feb 08 '15 at 07:39

pre-kidney

30,223

Would you like to give a sketch of the formal proof? – Yes Feb 08 '15 at 08:01
Yes, you just adapt the picture! First show that $|x^2-2x|-x+2$ has minimum at $3$ (by splitting into cases, and using the minimum of a quadratic function). This shows that when $k\leq -1$, there is at most 1 solution. All that remains is to consider $k\in [2,3]$. So we intersect with each of the two quadratics in turn, and find that they both have 2 solutions. You can use a continuity argument as well. – pre-kidney Feb 08 '15 at 08:01

score 1 · Answer 2 · answered Feb 08 '15 at 07:25

1

Hint: Graph $y_1 = |x^2-3x|$, and $y_2 = x+c$. Then see from the graphs what values of $c$ then $k-2$ yields two different intersections.

answered Feb 08 '15 at 07:25

DeepSea

77,651

Let $k$ be a real number. Prove that if the equation $|x^{2} - 3x| = x-2+k$ has two distinct roots, then either $-1 < k < 2$ or $k > 3$?

2 Answers2