Algebra with pq-formula

liljeros · #1 $Old$ May 22nd, 2008, 11:08 AM

Hello! I am currently having a homework for tomorrow with pretty advanced tasks, for me atleast, that I should solve. I have though got stuck on some of them but there is one in perticular that i have no idea how to solve and I was hoping that you might help me out with this. I start with writing the task.
(It is in Swedish so I am going to try to translate it the best i can)

a) Show that if an Quadratic equation has the roots $x_{1}=a x_{2}=b$ then the equation is $x^2-(a+b)x+ab=0$

b) Solve the Equation above and show that is has the solutions $x_{1}=a x_{2}=b$

I am quite sure that you need to use the pq-formula for this one but then you get $x=\frac{(a+b)}{2} \pm \sqrt{-(\frac{(a+b)}{2})^2}-ab$

This in my eyes does not say much.
If someone is willing to help me out here I'm very glad!
Thanks in advance!
//Rickard Liljeros

PS. I didn't put this in urgent help section since I maybe after this problem will ask for some other algebra help and thought it would be unnecessary to start a new thread for each one.

masters · #2 $Old$ May 22nd, 2008, 11:19 AM

If $x_1=a$ is a root, then $x-a$ is a factor.

If $x_2=b$ is a root, then $x-b$ is a factor.

Therefore,

$(x-a)(x-b)=x(x-b)-a(x-b)=x^2-bx-ax+ab=0$

Simplifying (factoring out x in 2nd and 3rd terms):

$x^2-(a+b)x+ab=0$

liljeros · #3 $Old$ May 22nd, 2008, 11:32 AM

Quote:

Originally Posted by masters $View Post$

If $x_1=a$ is a root, then $x-a$ is a factor.

If $x_2=b$ is a root, then $x-b$ is a factor.

Therefore,

$(x-a)(x-b)=x(x-b)-a(x-b)=x^2-bx-ax+ab=0$

Simplifying (factoring out x in 2nd and 3rd terms):

$x^2-(a+b)x+ab=0$

"If $x_1=a$ is a root, then $x-a$ is a factor.

If $x_2=b$ is a root, then $x-b$ is a factor."

How come it is like this? Is it a rule of some kind?

masters · #4 $Old$ May 22nd, 2008, 11:54 AM

You have probably learned that a zero of a function f(x) is any value c such that f(c)=0. When the function is graphed, the real zeros of the function are the x-intercepts of the graph.

If $f(x)=a_nx^n+...+a_1x+a_0$ is a polynomial function, then

c is a zero of the polynomial function f(x),

x-c is a factor of the polynomial function f(x), and

c is a root or solution of the polynomial equation f(x)=0.

liljeros · #5 $Old$ May 22nd, 2008, 12:06 PM

Quote:

Originally Posted by masters $View Post$

You have probably learned that a zero of a function f(x) is any value c such that f(c)=0. When the function is graphed, the real zeros of the function are the x-intercepts of the graph.

If $f(x)=a_nx^n+...+a_1x+a_0$ is a polynomial function, then

c is a zero of the polynomial function f(x),

x-c is a factor of the polynomial function f(x), and

c is a root or solution of the polynomial equation f(x)=0.

I have not learnd this yet, going on the first year in high school. But I think i maybe understand this.

masters · #6 $Old$ May 22nd, 2008, 12:26 PM

Quote:

Originally Posted by liljeros $View Post$

a) Show that if an Quadratic equation has the roots $x_{1}=a x_{2}=b$ then the equation is $x^2-(a+b)x+ab=0$

b) Solve the Equation above and show that is has the solutions $x_{1}=a x_{2}=b$

I am quite sure that you need to use the pq-formula for this one but then you get $x=\frac{(a+b)}{2} \pm \sqrt{-(\frac{(a+b)}{2})^2}-ab$

If you continue with your "pq" formula, you will arrive at the same conclusion; that $x_1=a \ and \ x_2=b$

This is a much longer process, but if that's the way you have to do it, then just continue with what you already started.

masters · #7 $Old$ May 22nd, 2008, 12:45 PM

$x=\frac{a+b}{2}\pm\sqrt{(\frac{a+b}{2})^2-ab}$
---------------------------------------------------------------
$x_1=\frac{a+b}{2} + \sqrt{\frac{a^2+2ab+b^2}{4}-ab}$

$x_1=\frac{a+b}{2}+\sqrt{\frac{a^2+2ab+b^2}{4}-\frac{4ab}{4}}$

$x_1=\frac{a+b}{2}+\sqrt{\frac{a^2+2ab+b^2-4ab}{4}}$

$x_1=\frac{a+b}{2}+\sqrt{\frac{a^2-2ab+b^2}{4}}$

$x_1=\frac{a+b}{2}+\frac{a-b}{2}$

$x_1=\frac{a+b+a-b}{2}$

$x_1=\frac{2a}{2}$

$x_1=a$

Now, perform a similar task to find $x_2$

liljeros · #8 $Old$ May 22nd, 2008, 12:57 PM

Quote:

Originally Posted by masters $View Post$

$x=\frac{a+b}{2}\pm\sqrt{(\frac{a+b}{2})^2-ab}$
---------------------------------------------------------------
$x_1=\frac{a+b}{2} + \sqrt{\frac{a^2+2ab+b^2}{4}}-ab$

$x_1=\frac{a+b}{2}+\sqrt{\frac{a^2+2ab+b^2}{4}}-\frac{4ab}{4}$

$x_1=\frac{a+b}{2}+\sqrt{\frac{a^2+2ab+b^2-4ab}{4}}$

$x_1=\frac{a+b}{2}+\sqrt{\frac{a^2-2ab+b^2}{4}}$

$x_1=\frac{a+b}{2}+\frac{a-b}{2}$

$x_1=\frac{a+b+a-b}{2}$

$x_1=\frac{2a}{2}$

$x_1=a$

Now, perform a similar task to find $x_2$

You really are the master!

A friend to me explained the thing you were talking about before and it seems like I have learnd that and it helped me alot! Thank you *2

Very kind of you to help me out!

topsquark · #9 $Old$ May 22nd, 2008, 01:11 PM

Quote:

Originally Posted by masters $View Post$

$x=\frac{a+b}{2}\pm\sqrt{(\frac{a+b}{2})^2-ab}$
---------------------------------------------------------------
$x_1=\frac{a+b}{2} + \sqrt{\frac{a^2+2ab+b^2}{4}}-ab$

$x_1=\frac{a+b}{2}+\sqrt{\frac{a^2+2ab+b^2}{4}}-\frac{4ab}{4}$

Just to be clear there is a typo in the last two of the first three lines of masters' post. (Probably because of the same typo in the original poster's statement.) They should read:
$x=\frac{a+b}{2}\pm\sqrt{(\frac{a+b}{2})^2-ab}$

$x_1=\frac{a+b}{2} + \sqrt{\frac{a^2+2ab+b^2}{4}-ab}$

$x_1=\frac{a+b}{2}+\sqrt{\frac{a^2+2ab+b^2}{4}-\frac{4ab}{4}}$

-Dan

masters · #10 $Old$ May 22nd, 2008, 01:16 PM

Thanks, Dan. I fixed it.