# Solve f (x) = 0

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• Oct 10th 2011, 05:45 AM
gilagila
Solve f (x) = 0
f (x) = 2 (x-3) (x+2)^2 + 2 (x+2) (x-3)^3

solve f(x) = 0

if we factorise, we get

f (x) = 2 (x-3) (x+2) (x^2 – 5x + 11)

0 = 2 (x-3) (x+2) (x^2 – 5x + 11)

is it meaning that x = 3, and x = 2 ?? or ?
• Oct 10th 2011, 05:49 AM
Prove It
Re: Solve f (x) = 0
Quote:

Originally Posted by gilagila
f (x) = 2 (x-3) (x+2)^2 + 2 (x+2) (x-3)^3

solve f(x) = 0

if we factorise, we get

f (x) = 2 (x-3) (x+2) (x^2 – 5x + 11)

0 = 2 (x-3) (x+2) (x^2 – 5x + 11)

is it meaning that x = 3, and x = 2 ?? or ?

Have you written this equation correctly? Is it \$\displaystyle \displaystyle f(x) = 2(x - 3)(x + 2)2 + 2(x + 2)(x - 3)3\$ or is it something else? Because if you factorise this, you should get \$\displaystyle \displaystyle f(x) = 2(x - 3)(x + 2)(2 + 3) = 2(x - 3)(x + 2)5 = 10(x - 3)(x + 2)\$...
• Oct 10th 2011, 05:52 AM
gilagila
Re: Solve f (x) = 0
Quote:

Originally Posted by Prove It
Have you written this equation correctly? Is it \$\displaystyle \displaystyle f(x) = 2(x - 3)(x + 2)2 + 2(x + 2)(x - 3)3\$ or is it something else? Because if you factorise this, you should get \$\displaystyle \displaystyle f(x) = 2(x - 3)(x + 2)(2 + 3) = 2(x - 3)(x + 2)5 = 10(x - 3)(x + 2)\$...

sorry , due to not sure how to use the quadratic and cubic sign, just change it to ^2 and ^3.
• Oct 10th 2011, 05:53 AM
HallsofIvy
Re: Solve f (x) = 0
One of the things I am sure yuo have already learned is that if abc= 0 then at least one of a, b, or c must be 0.
Yes, you can factor \$\displaystyle 2(x-3)(x+2)^2+ 2(x+2)(x-3)^3= 2(x+2)(x- 3)[(x+2)+ (x- 3)^2]= 2(x+2)(x-3)(x^2- 5x+ 11)= 0\$.

So you know immediately that x+2= 0, or x- 3= 0, or \$\displaystyle x^2- 5x+ 11= 0\$ is 0. If x- 3= 0, then x= 3 so x= 3 is a root. If x+ 2= 0 then x= -2 (NOT 2) so x= -2 is a root. But there are two other roots- numbers that satisfy \$\displaystyle x^2- 5x+ 11= 0\$. That does not have any real number roots but you can find the complex roots by using the quadratic formula.
• Oct 10th 2011, 05:59 AM
gilagila
Re: Solve f (x) = 0
Quote:

Originally Posted by HallsofIvy
One of the things I am sure yuo have already learned is that if abc= 0 then at least one of a, b, or c must be 0.
Yes, you can factor \$\displaystyle 2(x-3)(x+2)^2+ 2(x+2)(x-3)^3= 2(x+2)(x- 3)[(x+2)+ (x- 3)^2]= 2(x+2)(x-3)(x^2- 5x+ 11)= 0\$.

So you know immediately that x+2= 0, or x- 3= 0, or \$\displaystyle x^2- 5x+ 11= 0\$ is 0. If x- 3= 0, then x= 3 so x= 3 is a root. If x+ 2= 0 then x= -2 (NOT 2) so x= -2 is a root. But there are two other roots- numbers that satisfy \$\displaystyle x^2- 5x+ 11= 0\$. That does not have any real number roots but you can find the complex roots by using the quadratic formula.

Is it meaning that the x = 3, -2 and (5+√(-19))/2 or (5- √(-19))/2 <- using formula calculation ?
meaning if f(x) = 0, then x = { 3, 2, (5+√(-19))/2 and (5- √(-19))/2}
am i right ?
• Oct 10th 2011, 09:37 AM
SammyS
Re: Solve f (x) = 0
Quote:

Originally Posted by gilagila
Is it meaning that the x = 3, -2 and (5+√(-19))/2 or (5- √(-19))/2 <- using formula calculation ?
meaning if f(x) = 0, then x = { 3, 2, (5+√(-19))/2 and (5- &√(-19))/2}
am i right ?

Both x = (5+√(-19))/2 and x = (5- √(-19))/2 are solutions, although, as HallsofIvy mentioned, both of those are complex numbers. That's because \$\displaystyle x^2-5x+11>0\$ for all real number values of x. To see that, graph \$\displaystyle y=x^2-5x+11\,.\$
• Oct 10th 2011, 04:26 PM
gilagila
Re: Solve f (x) = 0
Quote:

Originally Posted by SammyS
Both x = (5+√(-19))/2 and x = (5- √(-19))/2 are solutions, although, as HallsofIvy mentioned, both of those are complex numbers. That's because \$\displaystyle x^2-5x+11>0\$ for all real number values of x. To see that, graph \$\displaystyle y=x^2-5x+11\,.\$

ok. Thanks for the reply