Thread: Solve f (x) = 0

1. 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 ?

2. Re: Solve f (x) = 0

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 $\displaystyle f(x) = 2(x - 3)(x + 2)2 + 2(x + 2)(x - 3)3$ or is it something else? Because if you factorise this, you should get $\displaystyle f(x) = 2(x - 3)(x + 2)(2 + 3) = 2(x - 3)(x + 2)5 = 10(x - 3)(x + 2)$...

3. Re: Solve f (x) = 0

Originally Posted by Prove It
Have you written this equation correctly? Is it $\displaystyle f(x) = 2(x - 3)(x + 2)2 + 2(x + 2)(x - 3)3$ or is it something else? Because if you factorise this, you should get $\displaystyle f(x) = 2(x - 3)(x + 2)(2 + 3) = 2(x - 3)(x + 2)5 = 10(x - 3)(x + 2)$...
sorry , due to not sure how to use the quadratic and cubic sign, just change it to ^2 and ^3.

4. 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 $2(x-3)(x+2)^2+ 2(x+2)(x-3)^3= 2(x+2)(x- 3)[(x+2)+ (x- 3)^2]= 2(x+2)(x-3)(x^2- 5x+ 11)= 0$.

So you know immediately that x+2= 0, or x- 3= 0, or $x^2- 5x+ 11= 0$ 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 $x^2- 5x+ 11= 0$. That does not have any real number roots but you can find the complex roots by using the quadratic formula.

5. Re: Solve f (x) = 0

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 $2(x-3)(x+2)^2+ 2(x+2)(x-3)^3= 2(x+2)(x- 3)[(x+2)+ (x- 3)^2]= 2(x+2)(x-3)(x^2- 5x+ 11)= 0$.

So you know immediately that x+2= 0, or x- 3= 0, or $x^2- 5x+ 11= 0$ 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 $x^2- 5x+ 11= 0$. 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 ?

6. Re: Solve f (x) = 0

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 $x^2-5x+11>0$ for all real number values of x. To see that, graph $y=x^2-5x+11\,.$

7. Re: Solve f (x) = 0

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 $x^2-5x+11>0$ for all real number values of x. To see that, graph $y=x^2-5x+11\,.$

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