Real Zeros of Quadratic Functions: x^2-3x-28, x^2+12x+36, x^2-3, 4-x^2

[Sonny18n]

Homework Statement

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

Homework Equations

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.

 

[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.

 

[Sonny18n]

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.

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

 

[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.

 

[Sonny18n]

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.

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.

 

[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 $x = \frac{-b\pm \sqrt{b^2 - 4ac}}{2a}$

 

[Sonny18n]

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 $x = \frac{-b\pm \sqrt{b^2 - 4ac}}{2a}$

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.

 

[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.

 

[Sonny18n]

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.

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?

 

[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 =$\sqrt{3}.\quad a^2 = x^2, b^2 = 3.$

 

[Mentallic]

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?

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.

 

[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 $x^2-3x- 28= (x- 7)(x+ 4)= 0 you must have x- 7= 0 and x+ 4= 0.$

 

[Ray Vickson]

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 $x^2-3x- 28= (x- 7)(x+ 4)= 0 you must have x- 7= 0 and x+ 4= 0.$

[itex] <br /> 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<b> OR</b> x+4 = 0.[/itex]

 

[HallsofIvy]

Yes, thank you.

 

[Sonny18n]

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 $x^2-3x- 28= (x- 7)(x+ 4)= 0 you must have x- 7= 0 and x+ 4= 0.$

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

 

[Ray Vickson]

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

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".

 

[Sonny18n]

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".

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

 

[Ray Vickson]

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

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

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 "=".

 

[Sonny18n]

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 "=".

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

 

[Mentallic]

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

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

 

[RUber]

You should try plotting the function. That might help you see where the zeros should be.