help with roots

iiharthero · #1 $Old$ July 4th, 2009, 12:11 AM

c) Find the value of p if (x-1)(x-2) = p has a double root
dont realli understand the qs at all or how to get to the answer

d) Show that if x=a is a double root of P(x)=0, then it is also a root of P'(x)=0
Let P(x)=(x-a)squaredQ(x)

thankyou for anyhelp !!!

Chris L T521 · #2 $Old$ July 4th, 2009, 12:19 AM

Quote:

Originally Posted by iiharthero $View Post$

c) Find the value of p if (x-1)(x-2) = p has a double root
dont realli understand the qs at all or how to get to the answer

d) Show that if x=a is a double root of P(x)=0, then it is also a root of P'(x)=0
Let P(x)=(x-a)squaredQ(x)

thankyou for anyhelp !!!

For part c), FOIL it out and see that $x^2-3x+2-p=0$

Now when we use the quadratic formula, the discriminant $\Delta$ tells us how many solutions we will have (real or complex).

If we have a double root, we want $\Delta=0\implies (-3)^2-4(1)(2-p)=0$ . From here, it follows that $9-8+4p=0\implies p=-\frac{1}{4}$ .

For part d), If $P\left(x\right)=\left(x-a\right)^2Q\left(x\right)$ , then $P\,^{\prime}\!\left(x\right)=2\left(x-a\right)Q\!\left(x\right)+\left(x-a\right)^2Q^{\prime}\!\left(x\right)$ . From here, its evident that $a$ is a root of $P\,^{\prime}\!\left(x\right)$ .

Does this make sense? If you have additional questions, feel free to post back!

iiharthero · #3 $Old$ July 4th, 2009, 02:50 AM

thankyou so much for your help, my first question makes complete sense now thankyou
but i dont'realli understand the second question ><""
uhm u used the product rule in differentation to get the equation, but why is a evidently a root of P'(x)?
thankyousomuch for ur time and effort =D

Prove It · #4 $Old$ July 4th, 2009, 02:54 AM

Quote:

Originally Posted by iiharthero $View Post$

c) Find the value of p if (x-1)(x-2) = p has a double root
dont realli understand the qs at all or how to get to the answer

d) Show that if x=a is a double root of P(x)=0, then it is also a root of P'(x)=0
Let P(x)=(x-a)squaredQ(x)

thankyou for anyhelp !!!

Yes, you use the product rule on $P(x)$ .

If $P(x) = (x - a)^2Q(x)$ , then

$P'(x) = (x - a)^2\frac{d}{dx}[Q(x)] + Q(x)\frac{d}{dx}[(x - a)^2]$

$= (x - a)^2Q'(x) + 2(x - a)Q(x)$

$= (x - a)[(x - a)Q'(x) + 2Q(x)]$ .

Since $(x - a)$ is a factor of $P(x)$ , that means $x = a$ is a root of $P(x)$ .