$x^2-2mx+m^2-1=0$ Find the range of m when one root lies in (-2, 4)

Question

Let: $x^2-2mx+m^2-1=0$. For all $m$ there exist real roots for the above equation. If one root lies in between $(-2,4)$ Find the value range for $m$

My Work

Since coefficient of $x^2$ is 1 the graph is a maximum. Also the discriminant is positive. How can I use this fact to solve this problem ? Please Help me. Thanks a lot

Answer 1

$$1=(x-m)^2\iff x=m\pm1$$

Check for

$2<m-1<4$

and $2<m+1<4$

Answer 2

The roots are $m+1$ or $m-1.$ Since one of them must stay within $(-2,4),$ we must have $-2<m+1<4,$ or $-2<m-1<4,$ which give $-3<m<3,$ or $-1<m<5.$ Thus, $m$ must satisfy the conditions $$-3<m<5.$$

Answer 3

$(x-m)^2-1=0;$

$(x-m-1)(x-m+1)=0;$

$x_1= m+1$; $x_2=m-1$;

1) $-2 <m+1<4; $

$-3<m<3;$

2) $-1 <m<5;$

Range: $(-3,3) \cup (-1,5)= (-3,5)$;

Refer to Bernhard's comment.

Fine tuning, find:

1) The range of m when exactly one root is in $(-2,4).$

2) The range of m when both roots are in $(-2,4).$

Answer 4

We have if your function is $f(x)$, then there must be a root in $[-2, 4]$ if $f(-2)f(4)<0$. (Why?)

So:

$$(m^2+4m+3)(m^2-8m+15)<0\to (m+1)(m+3)(m-3)(m-5)<0$$

Can you solve this?