• Oct 28th 2008, 07:30 AM
cougar
f(x) = x(squared) + kx + (k+3) where k is a constant. Given that the equation f(x) = 0 has qual roots..

A) Find the possible values of k
B) Solve f(x) = 0 for each possible value of k

Given instead that k = 8

C) Express f(x) in the form (x + a)squared + b, where a and b are constants.

• Oct 28th 2008, 07:42 AM
Jhevon
Quote:

Originally Posted by cougar
f(x) = x(squared) + kx + (k+3) where k is a constant. Given that the equation f(x) = 0 has qual roots..

should that be "equal roots"?

if so, then

Quote:

A) Find the possible values of k
B) Solve f(x) = 0 for each possible value of k
you need the discriminant to be zero

Quote:

Given instead that k = 8

C) Express f(x) in the form (x + a)squared + b, where a and b are constants.
this is asking you to complete the square, do you know how to do that? (you can google it :D)

Quote:

use the quadratic formula. do you remember it?
• Oct 28th 2008, 07:52 AM
Rafael Almeida
Sorry to ask, but I couldn't find anywhere what a 'qual root' is.

About (C): if k=8, then \$\displaystyle f(x) = x^2 + 8x + 24\$

Now you must use the fact that \$\displaystyle (a + b)^2 = a^2 + 2ab + b^2 \ (I)\$, which is a particular case of Newton's Binomial. If you examine the expression for f(x), you can see that taking a = x and b = 4 and applying it to (I) seems promising. In fact:
\$\displaystyle (x + 4)^2 = x^2 + 8x + 16\$
Notice that it's not f(x) yet, but it's close. All you need to do now is adding 8:
\$\displaystyle (x + 4)^2 + 8= x^2 + 8x + 16 + 8 = x^2 + 8x + 24 = f(x)\$
And this answers your questions, a = 4 and b = 8. The name of this procedure is Square Completion. Check the link for further examples.

For (D), I'll leave it to you with a tip: does what you just did in (C) helps you to find the roots? Or inform you whether they exist in \$\displaystyle \mathbb{R}\$?

Regards,
• Oct 28th 2008, 08:13 AM
cougar
Yeh oops a typo, supposed to be equal. Thanks for the help YOU RULE :D