Ok Geniuses: My kids algebra problem.

Gotta know just a little more information. What exactly do the instructions tell him to find? I'm assuming where the lines intersects?

 

Cool beans, give me a moment to knock my old brain around and I'll spit out an answer lol

 

Is this correct? You wrote 2 f(x)=......

Did you mean f(x)=3x^2+2x-4 ?

 

I'm thinking the answer is 4x^2 - 2x -3.

Let me explain why. Suppose x = 3.
Solve f(x) when x = 3. Then, solve g(x) when x = 3. Add those two answers together.
Compare that saved answer to solving (f+g)(x) when x = 3 and see if you get the same answer. Note that (f+g)(x) is the formula I provide above.

So, when x = 3, f(x) = 9 - 12 + 1 = -2. And, when x= 3, g(x) = 27 + 6 - 4 = 29. So, -2 + 29 = 27.
Now, when x = 3, (f + g)(x) = (4 * 9) - (2 * 3) - 3 = 36 - 6 - 3 = 27.

This should be true for any value of x.

So, the way you add any f(x) and g(x) formulas together is you keep each factor (that is, power of x) separate, adding together only "like powers of x".
In the above example, the x^2 "factor" is 3x^2 in f(x) and 1x^2 in g(x). Added together it is 4x^2.
And the x^1 "factor" is -4x in f(x) and 2x in g(x). Added together it is -2x.
And the x^0 "factor" is +1 in f(x) and -4 in g(x). Added together it is -3.
Putting the three "factor" parts together to make the new formula is 4x^2 - 2x - 3. This is (f+g)(x).

 

I didn't get anything near what your son got
3x^3 -10x^2 -8x + 1

 

I'm no expert, so please verify with someone who really knows math.

I am always completely logical and can defend my results - however, that isn't always the correct answer in school. YMMV.

 

Sir, this is an English speaking website and I will kindly ask you to refrain from posting in Chinese or Swahili or whatever that is. Thank you.

 

Da nada.

 

Ahhhhh...... pretty difficult to get a cubic (x^3) factor by adding squares (x^2) and linears (x), folks ----- in fact, it is impossible.

 

If I could add to what has already been posted.

Basically, (f+g)(x) = f(x) + g(x)

In this case both of these functions are parabolas. Imagine both of them graphed on the same chart. Each of them for a given x, will have a unique value of y. If you were to add the values of y for any given x you would have the same single value of y if you use the answer provided.

 

I'm inclined to agree with toggie

 

Correct, that's a better way of looking at it. So essentially..

(3x² + 2x -4) + (x² - 4x +1)

3x² - x² + 2x - 4x - 4 + 1

4x² - 2x - 3

 

That is correct.

"John's" (the fictitious boy in the original problem definition) mistake was that, for some reason, he has multiplied the correct answer by a factor of "x".

To clarify, let's call the correct solution h(x). Then, h(x) = 4x^2 - 2x - 3. Multiply h(x) by "x" as John did, and you will have --- 4x^3 - 2x^2 - 3x.

Isn't mathematics fun

 

And, hey Jim ----

Quit getting fellow F-chatters to do your son's homework for him!!!!!!

Just teasing, of course ----- happy to help

 

Ahhhhhhh........ that's "3x^2 (+) x^2....." I believe you made a typo there....

 

the correct answer is 4x^2-2x-3

johns mistake was he multiplied his answer by x

(f+g)x is a function that adds two individual functions, it doesn't mean f plus g times x