# Quadratic Equations with positive root

1. Aug 11, 2006

### konichiwa2x

How do you solve quadratic equations which have atleast one positive root??
Find the value of 'm' for which the roots are such that at least one is positive. x2 -(m-3)x + m = 0 mR.

Can someone help me get started?

Here is my work: I
checked the discriminant. for D>0,
m (-inf,-1) (9,inf)

2. Aug 11, 2006

### mathwonk

probably you use the fact that the sum of the roots is m-3 and the product of the roots is m.

3. Aug 11, 2006

### konichiwa2x

well how is that going to help? I have thought of all that.

4. Aug 12, 2006

### mathwonk

it seems either m < 0 or m >=9?

5. Aug 12, 2006

### d_leet

Should be

$$\frac{{m - 3 \pm \sqrt {m^2 - 10m + 9} }}{2}$$

If ax2 + bx + c = 0
then

$$x = \frac{{-b \pm \sqrt {b^2 - 4ac} }}{2a}$$

6. Aug 12, 2006

### bomba923

You have to find a value of 'm' such that at least one of the roots of x2-(m-3)x+m is positive.
Why not just use the quadratic formula?
$$x^2 - \left( {m - 3} \right)x + m = 0 \Rightarrow x = \frac{{{m - 3} \pm \sqrt {m^2 - 10m + 9} }}{2}$$

And so,
$$\begin{gathered} \frac{{m - 3 \pm \sqrt {m^2 - 10m + 9} }} {2} > 0 \Rightarrow \pm \sqrt {m^2 - 10m + 9} > 3 - m \Rightarrow \hfill \\ m^2 - 10m + 9 > \pm \left( {m^2 - 6m + 9} \right) \hfill \\ \end{gathered}$$

*For m2-10m+9 > m2-6m+9,
$$m^2 - 10m + 9 > m^2 - 6m + 9 \Rightarrow - 4m > 0 \Rightarrow m < 0$$

*For m2-10m+9 > -(m2-6m+9),
$$m^2 - 10m + 9 > - m^2 + 6m - 9 \Rightarrow m^2 - 8m + 9 > 0 \Rightarrow m > 4 + \sqrt 7 \text{ or } m < 4 - \sqrt 7$$

However, x2-(m-3)x+m has a real discriminant only for m≤1 or m≥9.

The inequality m2-10m+9 > m2-6m+9 derived m<4-sqrt(7) and m>4-sqrt(7), the former of which is weaker than m<0 (as derived from the inequality m2-10m+9 > m2-6m+9), and the latter of which is weaker than the condition m≥9 (necessary for x2-(m-3)x+m to have real discrimants when m≥0).

Thus, choose an m<0 or an m≥9 and the problem is solved.

7. Aug 12, 2006

### VietDao29

Are you saying that $$\begin{gathered} \pm \sqrt{A} > B \\ \Rightarrow A > \pm B ^ 2 \end{gathered}$$?
Say, we have to solve the equation:
$$\begin{gathered} \pm \sqrt{x} > x + 1 \\ \Rightarrow x > \pm (x ^ 2 + 2x + 1) \end{gathered}$$
** x > x2 + 2x + 1 $$\Rightarrow x \in \emptyset$$
** x > -x2 - 2x - 1 $$\Rightarrow \left[ \begin{array}{l} x < \frac{-3 - \sqrt{5}}{2} \\ x > \frac{-3 + \sqrt{5}}{2} \\ \end{array} \right.$$
And $$\sqrt{x}$$ is defined for x >= 0, and $$\frac{-3 + \sqrt{5}}{2} < 0 \leq x$$, so the solution to the inequality is x >= 0???
Say, x = 0.5, is:
$$\pm \sqrt{0.5} > 0.5 + 1 = 1.5$$?

Last edited: Aug 12, 2006
8. Aug 12, 2006

### VietDao29

Say you have two roots x1 < x2
** Say, the larger root of the equation is non-positive, i.e x2 <= 0. So we'll have:
x1 < x2 < 0. Both roots are non-positive. And this is not the case we are looking for, right?
** If the larger root of the equation is positive, then of course we have at least one positive root, namely x2.
So what we will have to do is to find the cases in which the larger root is positive. Right?
There are two roots of the equation above:
$$x = \frac{m - 3 \pm \sqrt{m ^ 2 - 10 m + 9}}{2}$$, which is the larger one?
Can you get me?
Can you go from here? :)

Last edited: Aug 12, 2006
9. Aug 12, 2006

### konichiwa2x

yup thanks for the help all.