Values in Quadratic Equation with 2 different roots

FlopperJr · Nov 27, 2011

1. The problem statement, all variables and given/known data

Find the intervals of all possible value of p which the equation equation: (p-1)x^2+4x+(p-4)=0 has two different roots.

2. Relevant equations

ax^2+bx+c>0 ??

3. The attempt at a solution

(p-1)x^2+4x+(p-4)>0 ??

How would I go about solving this?
Is two roots >0 or =0??

Could I have an example so that I could then solve the problem on my own or steps I need to take in order to solve for the values of p.

ehild · Nov 27, 2011

Can you solve a quadratic equation? What formula do you use for the solution? In case of p=2 you have the quadratic equation x^2+4x-2. What are the roots?

ehild

FlopperJr · Nov 27, 2011

yes but I dont know with (p-1) and (p+4)

-b+/- Squareroot b^2 - 4ac over 2a ?

Do i plug it in that?

Mark44 · Nov 27, 2011

Yes. Use the quadratic formula with a = p - 1, b = 4, and c = p - 4.

FlopperJr · Nov 27, 2011

ohh okay. I used a pgraphing calculator and think I got the answer.
p< or equal to 4

and
p>1

ehild · Nov 27, 2011

Are you sure? What are the roots if p=0.5? When p=4.5?

ehild

FlopperJr · Nov 27, 2011

Ohh. I see. I believe my answer now is p<5 and p>0.

ehild · Nov 27, 2011

It is better. But why do you think so?
When has a quadratic equation two different roots?

ehild

FlopperJr · Nov 27, 2011

When it intersects the x axis. It intersects at two different points. So I just plugged in values for p until I began to see two different roots and then narrowed down my answer.

ehild · Nov 28, 2011

FlopperJr said: ↑

When it intersects the x axis. It intersects at two different points. So I just plugged in values for p until I began to see two different roots and then narrowed down my answer.

Yu can not get the exact solution from a graph. By the way, people could (and can) solve quadratic equations without graphing calculators.
Use the formula that you copied already for the solution of the quadratic equation

ax²+bx+c=0

[tex]x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]

Because of the ± in front of the square root, the quadratic equation has two real roots if the expression under the square root (the discriminant) is greater than zero. There is a single solution if it is equal to zero, and no real solution if the discriminant is negative.
So you need to find those p values which make the discriminant greater than zero:
[tex]b^2-4ac>0[/tex]
Plug in the expressions for a, b, and c, and solve the inequality for p.

ehild

FlopperJr · Nov 28, 2011

Thank you so much for your help!!!! i could not have solved it without your help. I get it. Now lets just see how do on my test this Wednesday.

ehild · Nov 28, 2011

Good luck!

ehild

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Values in Quadratic Equation with 2 different roots

FlopperJr

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FlopperJr

Mark44

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