Finding zeroes of a polynomial with parameter

Question

Problem description says that for the function:

$$ f(x) = mx^2 + x + m - 1 $$

There are two different zeroes out of which both are lesser than 1. I'm tasked with finding all values of m that would fulfill this requirement.

First step was to see that the function is linear if the m parameter is 0, so the first m value that fulfills the requirement is the number 0. Then:

$$ b^2 - 4ac = 1 - 4m * (m-1)$$ $$ b^2 - 4ac = -4m^2 + 4m + 1$$

Then calculating $$ b^2 - 4ac $$ with respect to the above, like this:

$$ 16 - 4 * (-4) * 1 = 0 $$ $$ m_0 = \frac{-4}{-8} = \frac{1}{2} $$

So my final result is:

$$ m \in (0, \frac{1}{2} ) $$

But the answer sheet says that the actual answer is:

$$ m \in (0, \frac{1 + \sqrt2}{2} ) $$

So what did I do wrong here?

It's unclear what you did after you calculated the discriminant. What do you mean by "calculating $ b^2 - 4ac $ with respect to the above"? — Adam Latosiński, Sep 11 '19 at 14:48
M One, this is not a correct result ($m=1/2$ isn't the zero of the discriminant), and two, why would you think this solves the problem? — Adam Latosiński, Sep 11 '19 at 14:55
I'm not really sure why you did what you did. But I would start with calculating the roots in terms of m. — Erin, Sep 11 '19 at 14:56

score 1 · Answer 1 · answered Sep 11 '19 at 14:57

It looks like you are following some algorithm that used to solve a similar problem, but it doesn't make much sense in this problem

The function must have

Two roots (not one, not zero)

$ax^2+bx+c$ has two roots only when $a\neq0$ (linear function has only one root) and the $D>0$.

Each root should be less than 1.

Since it's easy to say which root is the largest, we can check that only largest root is smaller than 1.

Each of those conditions will give us sets of possible values of $m$. We should take the intersection of those sets, since we need those conditions to be true at the same time.

Can you figure out the rest?

Z Ahmed · Answer 2 · 2019-09-11T15:34:54.017

Let $$f(x)=mx^2+x+m-1.$$ $f(x)=0$ will have real roots if $$ B^2\ge 4AC \Rightarrow 4m^2-4m-1 \le 0 \Rightarrow \frac{1-\sqrt{2}}{2} \le m \le \frac{1+\sqrt{2}}{2} ----(1),$$ and if $$-B/2A <1 \Rightarrow \frac{-1}{2m}<1 \Rightarrow m(2m+1)>0 \Rightarrow m>0~ {\mbox or}~ m<-1/2-----(2),$$ and if $$f(1)>0 \Rightarrow 2m>0......(3)$$ The overlap of (1,2,3) gives $$ m \in (0,\frac{1+\sqrt{2}}{2}),$$ for two real distinct roots <1.

score 0 · Accepted Answer · answered Sep 11 '19 at 14:57

The method you tried to solve the problem isn't clear to me, so I can't tell what are you doing wrong. But to solve the problem, I suggest using Viete's formulas.

The condition "two different zeros, both lesser than 1" can be expressed by conditions. $$ \Delta > 0 \\ (x_1-1) + (x_2-1) < 0 \\ (x_1-1)\cdot (x_2-1) > 0$$ that is $$ \Delta > 0 \\ x_1+x_2-2 < 0 \\ x_1x_2 - (x_1+x_2) +1 > 0$$ You have $\Delta= b^2-4ac=-4m^2+4m+1$ and the Viete's formulas give $x_1+x_2 = \frac{b}{a} = \frac{1}{m}$ and $x_1x_2 = \frac{c}{a}=\frac{m-1}{m}$. Therefore you need to solve the set of equations: $$ -4m^2+4m+1 > 0 \\ \frac{1}{m}-2 < 0 \\ \frac{m-1}{m} - \frac{1}{m} +1 > 0$$

Finding zeroes of a polynomial with parameter

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