# Finding Real Zeros

1. Mar 12, 2015

### Sonny18n

1. The problem statement, all variables and given/known data
Find the real zeros of the following quadratic functions. Every problem will have an f(x)= in front of it.
1) x^2- 3x- 28
2) x^2+12x+36
3)x^2- 3
4) 4- x^2
2. Relevant equations

3. The attempt at a solution
1) Common factors are 7 and 4.
So would it be (x+7)?(x-4)=0?
Which means the zeros would be -7 and 4. If I'm doing it right that is.

Last edited by a moderator: Mar 12, 2015
2. Mar 12, 2015

### RUber

I think you are on the right track.
For the first one, you correctly identified the 4 and 7, but your signs were wrong when you put them into the polynomials.
(x+7)(x-4) = x^2 +7x -4x -28 = x^2 +3x -28. Switch the signs, and you should be good to keep going in that direction to solve the rest.

3. Mar 12, 2015

### Sonny18n

Wait so the zeros are 7 and -4? I saw somewhere that the factoring and zeros were opposite. Or am I confusing myself?

4. Mar 12, 2015

### RUber

You may be confusing the hint.
Your factored form of (1) should be (x-7)(x+4) = x^2 -7x+4x-28 = x^2-3x-28 which is what you posted.
The zeros of this are clearly when (x-7)=0 or when (x+4)=0. This happens when x = 7 or -4.
Just saying signs are opposite sometimes leads to switching signs where you shouldn't just because you think you are supposed to change them somewhere.

5. Mar 12, 2015

### Sonny18n

But are there cases where that happens because I get frustrated at that concept. Anyways
2) The 2 zeros would both be positive 6, right?
As for 3 and 4, I just need an example how to factor functions that look like that.

6. Mar 12, 2015

### RUber

for (2) your factors are (x+6)^2, is that zero when x=6? Finding zeros mean find an x such that the function equals zero at that x.
For 3 and 4, I recommend you use the difference of squares formula:
$a^2 - b^2 = (a-b)(a+b)$
When in doubt, the quadratic equation will always work:
For $f(x) = ax^2 + bx +c$ zeros are at $x = \frac{-b\pm \sqrt{b^2 - 4ac}}{2a}$

7. Mar 12, 2015

### Sonny18n

For 2 I skipped to the end so I can cover more ground at a quicker pace, Found the common factors and matched with positive 12.

I'm still confused on solving 3 and 4. a^2 - b^2 seems simple enough but look at 3. a should be the first number but it's x^2.

8. Mar 12, 2015

### RUber

For (2) you found the right factors, x+6 and x+6. The zeros are not at +6.
For (3) you have x^2 - 3. For this form, you want a^2 = x^2 and b^2 = 3.

9. Mar 13, 2015

### Sonny18n

Sorry but I still don't understand. What am I doing wrong in 2? And how do I get those functions to equal x^2 and 3?

10. Mar 13, 2015

### RUber

If you are trying to find the zeros from a factored form, you need to make the factors zero.
2 factors to (x+6)(x+6), what do you get for x=6? 6^2 + 12*6+ 36 = 144. You need to change the sign -- x=-6 gives (-6)^2 +12(-6)+ 36 = 72-72 = 0.
This should make sense since (x+6) is zero when x=-6, and anything times zero is zero.

For 3, let x = a and b =$\sqrt{3}.\quad a^2 = x^2, b^2 = 3.$

11. Mar 13, 2015

### Mentallic

Since you're obviously trying to learn this quickly, I'll give a short crash course on factorizing.

We have the quadratic $f(x)=x^2+12x+36$.

When we talk about f(x), we are actually talking about $x^2+12x+36$ in this case.
f(1) will then represent f(x) when x=1, hence
$$f(x)=x^2+12x+36$$
and
$$f(1)=(1)^2+12(1)+36=1+12+36=49$$

Similarly, $f(6)=(6)^2+12(6)+36=12+72+36=144$ and $f(-6)=(-6)^2+12(-6)+36=36-72+36=0$

Notice f(-6)=0, hence x = -6 is a zero of the quadratic f(x). But of course, it's not easy to figure out the values of x this way. We need to turn to factorizing.
Since you already know how to factorize, we'll move on.

So $f(x)=x^2+12x+36=(x+6)^2$.

Now, f(1)=49 that we saw earlier, but with this factorized form, we can find that out a lot more easily. $f(x)=(x+6)^2$ hence $f(1)=(1+6)^2=7^2=49$.
Similarly,
$$f(6)=(6+6)^2=12^2=144$$
and
$$f(-6)=(-6+6)^2=0^2=0$$

It was easy to see that x=-6 gave us f(x)=0 because if k represents any number then k2=0 only when k=0. So if we think of x+6 as being k, hence $(x+6)^2=k^2$ then we can only get $(x+6)^2=0$ by letting x+6=0, solving gives us x=-6.

12. Mar 13, 2015

### HallsofIvy

iYou need to stop memorizing formulas and think about what you are doing! To "find a zero of f(x)" means to solve the equation f(x)= 0. So to find a zero 0 of $f(x)= x^2- 3$ means to solve the equation $x^2- 3= 0$. I presume you know that you can add the same thing to both sides of the equation so that is the same as $x^2= 3$. There are two numbers that make that true.

You should also know that if ab= 0 then either a= 0 or b= 0 or both. So to solve [itex]x^2-3x- 28= (x- 7)(x+ 4)= 0 you must have x- 7= 0 and x+ 4= 0.

Last edited by a moderator: Mar 13, 2015
13. Mar 13, 2015

### Ray Vickson

To not confuse the OP even more, that should say: to solve $x^2 - 3x - 28 = (x-7)(x+4) = 0$ you must have x-7 = 0 OR x+4 = 0.

14. Mar 13, 2015

### HallsofIvy

Yes, thank you.

15. Mar 15, 2015

### Sonny18n

I think I understand how you got to the x^2=3 part but is the way you get the zero by squaring the 3. You said there were two numbers and.. Ugh

16. Mar 15, 2015

### Ray Vickson

No, you do NOT square the 3---you do the exact opposite! Of course, there are two roots; if x = r is one root, so is x = -r, because $(-r)^2 = (+r)^2$, so both -r and +r have the same "square".

17. Mar 15, 2015

### Sonny18n

Crap, you're right, I meant square root. So the answer is 1.73?

18. Mar 15, 2015

### Ray Vickson

No, because $\sqrt{3} \neq 1.73$. Your 1.73 is an approximation, not the exact value. In fact, you cannot write down any finite decimal number that gives the answer, because $\sqrt{3}$ is a so-called irrational number. A better approximation to the answer is $\sqrt{3} \doteq 1.732050808$; an even better approximation is $\sqrt{3} \doteq 1.7320508075688772935274463415058723669428052538104$, but that is still not exact. (Note that I used "$\doteq$" instead of "$=$", to indicate these are approximations.)

Of course, in practice you might use a calculator to compute an answer, in which case you would make do with a decimal approximation such as 1.73 or 1.7321 or whatever meets the accuracy of your analysis. However, I am trying to get you to realize that you would be dealing with an approximate answer, and somewhere you should indicate that you understand that---either by saying "approximately", or by using a symbol such as "≈" instead "=".

Last edited: Mar 15, 2015
19. Mar 16, 2015

### Sonny18n

Got it, thanks. Can you help me understand 2 a little better?

20. Mar 16, 2015

### Mentallic

Was my post at #11 not sufficient? What don't you understand?