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The question says-if roots of $x^2 -2mx +m^2 -1$ lie in $(-2,4)$, Prove that $m$ lies in $(-1,3)$. This was pretty easy, but then I tried to work backwards. What if the question read-"If $m$ lies between $(-1,3)$, prove that the roots of $x^2 -2mx +m^2 -1$ lie in $(-2,4)$". I tried a lot, but I really don't know how to proceed. Can anyone help me out?(This is not a homework problem, though!)

Thanks in advance!!

EDIT- The 'forward' method.

Suppose the roots are the $\alpha$ and $\beta$. It is given that they lie in $(-2,4)$. Let the given expression be $f(x)$. It is easy to see that $f(-2)$ and $f(4)$ are both greater than $0$.( you can draw the X axis, draw an upward parabola(since a>0 here) intersecting it at $(\alpha,0)$ and $(\beta,0)$. As they lie between -2 and 4, mark (-2,0) and (4,0) on either side of the parabola's intersections. Draw perpendiculars from these points to the parabola. You will see that both $f(-2)$ and $f(4)$ are positive.) Now, expand out $f(-2)$ and $f(4)$, and put them both $>0$. Solve the inequalities.

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  • $\begingroup$ Please share your "pretty easy" forward method! $\endgroup$ – lab bhattacharjee Sep 19 '15 at 6:25
  • $\begingroup$ there you go @labbhattacharjee $\endgroup$ – GRrocks Sep 19 '15 at 6:45
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Now Given $\displaystyle x^2-2mx+m^2-1 = 0\Rightarrow (x-m)^2 = 1\Rightarrow x=\pm 1+m$

So we get $m=x\mp 1\;,$ Given $-1<m<3$

$\bullet \;$ If $ m=x-1\;,$ Then $-1<x-1<3\Rightarrow 0<x<4$

$\bullet \;$ If $ m=x+1\;,$ Then $-1<x+1<3\Rightarrow -2<x<2$

so we have seen above $x$ lies between $-2$ to $4$

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  • $\begingroup$ Welcome GRrocks. $\endgroup$ – juantheron Sep 19 '15 at 6:24
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HINT:

$$x=\dfrac{2m\pm\sqrt{(2m)^2-4(m^2-1)}}2=m\pm1$$

Now use $-1<m<3$ to find the ranges of $m\pm1$

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  • $\begingroup$ I was tempted to start with $$-1<m<3\iff-2<m-1<2\iff-1<\dfrac{m-1}2<1,$$ Let $\dfrac{m-1}2=\cos A\iff m=2\cos A+1$ which is not necessary $\endgroup$ – lab bhattacharjee Sep 19 '15 at 6:21

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