Quadratic Equations with positive root

[konichiwa2x]

How do you solve quadratic equations which have atleast one positive root??
The question was this:
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)
Now what do i do? please help

 

[mathwonk]

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

 

[konichiwa2x]

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

 

[mathwonk]

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

 

[d_leet]

$$x^2 - \left( {m - 3} \right)x + m = 0 \Rightarrow x = \frac{{3 - m \pm \sqrt {m^2 - 10m + 9} }}{2}$$

Should be

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

Since the quadratic formula says
If ax² + bx + c = 0
then

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

 

[bomba923]

How do you solve quadratic equations which have atleast one positive root??
The question was this:
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)
Now what do i do? please help

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

*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.

 

[VietDao29]

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

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:
$$\pm \sqrt{0.5} > 0.5 + 1 = 1.5$$?

 

[VietDao29]

How do you solve quadratic equations which have atleast one positive root??
The question was this:
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)
Now what do i do? please help

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? :)

 

[konichiwa2x]

yup thanks for the help all.