# parabola problems, and other questions

• Jan 29th 2010, 01:01 PM
koumori
parabola problems, and other questions
Hello, I have a few questions that are making my head hurt.

Question 1:
if the parabola $\displaystyle y -2=a(x-3)^2$ goes through the point (2,0), what is the value of a?

Question 2:
Given $\displaystyle x^3-4x^2+2x+1=0$
a:How many possible positive roots are there?
b:How many possible negative roots are there?
c:What are the possible negative roots?
d:Using synthetic substitution, which of the possible rational roots is actually a root of the equation?
e:Find the irrational roots of the equation.(Hint:Use the quadratic formula to solve the depressed equation.)

Thanks
• Jan 29th 2010, 01:28 PM
Henryt999
First one..
Given $\displaystyle y-2 = a(x-3)^2$ /devide both sides by $\displaystyle (x-3)^2$

then $\displaystyle a = (y-2)/(x-3)^2$ you know that when x = 2; y = 0 so just plug those in.. and there you have it...
• Jan 29th 2010, 01:29 PM
pickslides
Quote:

Originally Posted by koumori
Question 1:
if the parabola $\displaystyle y -2=a(x-3)^2$ goes through the4 point (2,0), what is the value of a?

Thanks

$\displaystyle 0 -2=a(2-3)^2$

now solve for $\displaystyle a$ . You may need to use complex numbers.

Quote:

Originally Posted by koumori

Question 2:
Given $\displaystyle x^3-4x^2+2x+1=0$
a:How many possible positive roots are there?
b:How many possible negative roots are there?
c:What are the possible negative roots?
d:Using synthetic substitution, which of the possible rational roots is actually a root of the equation?
e:Find the irrational roots of the equation.(Hint:Use the quadratic formula to solve the depressed equation.)

Thanks

This one is a little harder, what have you tried?
• Jan 29th 2010, 01:39 PM
Henryt999
how?
Just curious, how could a complex number be involved?

I just get $\displaystyle a = -2$?

what am I missing?
• Jan 29th 2010, 02:18 PM
BabyMilo
Quote:

Originally Posted by koumori
Hello, I have a few questions that are making my head hurt.

Question 1:
if the parabola $\displaystyle y -2=a(x-3)^2$ goes through the point (2,0), what is the value of a?

Question 2:
Given $\displaystyle x^3-4x^2+2x+1=0$
a:How many possible positive roots are there?
b:How many possible negative roots are there?
c:What are the possible negative roots?
d:Using synthetic substitution, which of the possible rational roots is actually a root of the equation?
e:Find the irrational roots of the equation.(Hint:Use the quadratic formula to solve the depressed equation.)

Thanks

2) you know something times something = 1
so you try 1 for example,
$\displaystyle 1^3-4(1)^2+2(1)+1=0$

so (x-1) is one of the root.

divide the equation by $\displaystyle (x-1)$
$\displaystyle (x-1)(x^2-3x-1)=0$

use the quadratic equation. to solve for the other two roots.

2a) 2
b)1
e)$\displaystyle \frac{3+/-\sqrt{(-3)^2-4*1*-1)}}{2}$
$\displaystyle =\frac{3+/-\sqrt{13}}{2}$

hope this helps.
• Jan 29th 2010, 02:40 PM
koumori
Quote:

Originally Posted by pickslides
Quote:

Originally Posted by koumori
Question 2:
Given $\displaystyle x^3-4x^2+2x+1=0$
a:How many possible positive roots are there?
b:How many possible negative roots are there?
c:What are the possible negative roots?
d:Using synthetic substitution, which of the possible rational roots is actually a root of the equation?
e:Find the irrational roots of the equation.(Hint:Use the quadratic formula to solve the depressed equation.)

This one is a little harder, what have you tried?

Well...I'm pretty sure I know the answers for a,b, and c: (a:1)(b:1)(c:-\frac{1}{1} , \frac{1}{1}.
For "d" I've used synthetic substitution and I think that the answer is -1, that last one is the one that makes my head hurt.
If I'm not much mistaken, the depressed equation is: [tex]x^2+3x+1=0[\math]
when I use the quadratic equation on it, I get a gibberish answer that is rational:-2.168 approximately.
• Jan 30th 2010, 01:19 AM
mr fantastic
Quote:

Originally Posted by koumori
Well...I'm pretty sure I know the answers for a,b, and c: (a:1)(b:1)(c:-\frac{1}{1} , \frac{1}{1}.
For "d" I've used synthetic substitution and I think that the answer is -1, that last one is the one that makes my head hurt.
If I'm not much mistaken, the depressed equation is: [tex]x^2+3x+1=0[\math]
when I use the quadratic equation on it, I get a gibberish answer that is rational:-2.168 approximately.

x = 1 is clearly a solution. Therefore x - 1 is a factor of $\displaystyle x^3 - 4x^2 + 2x + 1$. Therefore the quadratic factor is $\displaystyle x^2 - 3x - 1$.
• Feb 2nd 2010, 08:13 AM
koumori
ok, I think I really went wrong somewhere!
When I do synthetic division on the equation I find that -1 gives a remainder of 0....
I remember that if there is no remainder left after doing synthetic division then the divisor is a factor of the equation.

Now I am confused.....where did I mess up?
• Feb 3rd 2010, 12:35 AM
mr fantastic
Quote:

Originally Posted by koumori
ok, I think I really went wrong somewhere!
When I do synthetic division on the equation I find that -1 gives a remainder of 0....
I remember that if there is no remainder left after doing synthetic division then the divisor is a factor of the equation.

Now I am confused.....where did I mess up?

Since you have not posted your working it's impossible to know where you messed up (and yes, you have messed up somewhere because (-1)^3 - 4(-1)^2 + 2(-1) + 1 does not equal zero).
• Feb 4th 2010, 03:57 PM
koumori
Ok....lesson learned!
1 is the number to use, so that means that the depressed equation is:
$\displaystyle x^2 - 3x - 1$ then, the result, when the quadratic equation is applied is 3+i or 3-i?
• Feb 4th 2010, 06:15 PM
Masterthief1324
Quote:

Originally Posted by pickslides
$\displaystyle 0 -2=a(2-3)^2$

now solve for $\displaystyle a$ . You may need to use complex numbers.

This one is a little harder, what have you tried?

Complex numbers are unnecessary to solve the equation you posted. -1 squared is 1. Therefore you have -2 = a*1; thus $\displaystyle a = \frac {-1}{2}.$
• Feb 4th 2010, 08:21 PM
mattio
Try using Descartes' Rule of Signs
• Feb 5th 2010, 09:18 AM
koumori
Quote:

Originally Posted by mattio
Try using Descartes' Rule of Signs

Thanks! I did, I got an answer of 1 possible positive root and one possible negative root.
Is that correct?