Prove One Root in Interval (0,1) for Equation with Real Numbers

[anemone]

Here is this week's POTW:

-----

Let $m,\,n,\,k$ be real numbers such that $m>0$ and $\dfrac{m}{5}+\dfrac{n}{4}+\dfrac{k}{3}=0$.

Prove that the equation $mx^2+nx+k=0$ has one root in the interval $(0,\,1)$.

-----

Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[anemone]

Congratulations to greg1313 for his correct solution:).

greg1313's solution:

Spoiler

\(\displaystyle \frac m5+\frac n4+\frac k3=0\)

Divide both sides by \(\displaystyle -m\):

\(\displaystyle -\frac 15-\frac{n}{4m}-\frac{k}{3m}=0\)

Using Vieta's formulae for the sum and product of the roots we have

\(\displaystyle -\frac 15+\frac{a+b}{4}-\frac{ab}{3}=0\)

\(\displaystyle 3(a+b)-4ab=\frac{12}{5}\Rightarrow b=\frac{12-15a}{15-20a},\quad0<\frac{12-15a}{15-20a}<1,\quad\forall\,\, a\in\mathbb{R},a\cancel\in(0,1)\)

To show 0 < b < 1 over the intervals \(\displaystyle (-\infty,0]\) and \(\displaystyle [1,\infty)\),

\(\displaystyle f(a)=\frac{12-15a}{15-20a},\quad f'(a)=\frac{-15(15-20a)+20(12-15a)}{(15-20a)^2}=\frac{15}{(15-20a)^2}\)

As its derivative is always positive over the given intervals, \(\displaystyle f(a)\) is strictly increasing over the given intervals.

Now, \(\displaystyle \lim_{a\to-\infty}f(a)=\frac34\) and \(\displaystyle f(0)=\frac45\) so \(\displaystyle 0<b<1\) on \(\displaystyle (-\infty,0]\)

\(\displaystyle \lim_{a\to\infty}f(a)=\frac34\) and \(\displaystyle f(1)=\frac35\) so \(\displaystyle 0<b<1\) on \(\displaystyle [1,\infty)\)

as required.

I will note that if \(\displaystyle a\in(0,1)\) and \(\displaystyle a\ne\frac34\) we are done.

You can also find the proposed solution below:

Spoiler

Let $f(x)=mx^2+nx+k$.

Note that we're given $\dfrac{m}{5}+\dfrac{n}{4}+\dfrac{k}{3}=0$, multiply both sides by the quantity $5\left(\dfrac{4}{5}\right)^2$, we get:

$5\left(\dfrac{4}{5}\right)^2\left(\dfrac{m}{5}+\dfrac{n}{4}+\dfrac{k}{3}\right)=5\left(\dfrac{4}{5}\right)^2(0)$

$\left(\dfrac{4}{5}\right)^2m+\dfrac{4}{5}n+\dfrac{4^2}{3\cdot 5}k=0$

$\left(\dfrac{4}{5}\right)^2m+\dfrac{4}{5}n=-\dfrac{16}{15}k$

$\left(\dfrac{4}{5}\right)^2m+\dfrac{4}{5}n+k=-\dfrac{k}{15}$

In other words, we have

$f\left(\dfrac{4}{5}\right)=\left(\dfrac{4}{5}\right)^2m+\dfrac{4}{5}n+k=-\dfrac{k}{15}$ and $f(0)=m(0)^2+n(0)+k=k$

Since $f$ is a continuous function over all real $x$, and $f(0)f\left(\dfrac{4}{5}\right)=k\left(-\dfrac{1}{15}k\right)=-\dfrac{k^2}{15}\le 0$, Intermediate value theorem tells us that there is a root in the interval $\left(0,\,\dfrac{4}{5}\right]\subset (0,\,1)$, and we're done.