Find the zeros by using the quadratic formula
f(x)=x^2+8x-3

Question

Find the zeros by using the quadratic formula
f(x)=x^2+8x-3

debbieg · Answer

A quadratic is an function of the form $\Large f(x)=ax^2+bx+c$

The quadratic formula gives you the zeros of such a function (that is, it gives you the x values that are the solutions to $\Large f(x)=0$) according to:

$$\Large x=\frac{ -b \pm \sqrt{b^2-4ac} }{ 2a }$$

(if you don't know that, you should probably work on memorizing it, it comes up a lot :)

So you just need to look at your function, figure out what a, b and c a are equal to (can you tell me?).

Then plug & chug in the formula! Those are your zeros.

anonymous · Answer

so far i'm at $$x=\pm \sqrt{(8)^{2}-4(1)(3)}$$ all over 2(1)

debbieg · Answer

Ok, that's not TOO far off, but a couple of problems.

first of all, you are missing something in the numerator. before the $\pm$, there is a -b, so don't forget about that.

Under the radical, you need to be very careful with your signs.  Take another look at "c", and see if you can fix the part under the radical sign. :)

anonymous · Answer

i meant to put -8 before the plus-minus.

anonymous · Answer

okay , i have $$x=-8\pm \sqrt{76}$$

debbieg · Answer

Good! Of course, that's over the den'r of 2.

Now you need to simplify the radical part. Do you remember how to simplify $\sqrt{76}$?

HINT: you need to pull out any FACTORS of 76 that are PERFECT SQUARES. :)

anonymous · Answer

2 and 38?

debbieg · Answer

Hmm. Well, those are two factors of 76, but that doesn't help you since neither of those are perfect squares.

but you can factor 76 further.... right? Do the complete prime factorization, and it should be obvious what factors (if any) are perfect squares.

anonymous · Answer

4 and 19?

debbieg · Answer

Ok, now one of those is a perfect square, right? Do you know now how to simplify $\sqrt{76}$?

debbieg · Answer

You know that $\sqrt{76}=\sqrt{4\cdot 19}$

How do you simplify that?

anonymous · Answer

wouldn't the answer be $$x=-4\pm2\sqrt{19}$$ ?

debbieg · Answer

Remember:

$\Large  \sqrt{m^2}=m $
and
$\Large  \sqrt{m^2\cdot n }=m\sqrt{n } $

debbieg · Answer

Well, that's VERY close.

What did you get for the simplified form of $\sqrt{76}$?

anonymous · Answer

$$x=-8\pm \sqrt{4}\sqrt{19}$$ ?

debbieg · Answer

Ok, i'm not sure what you're telling me there. I'm asking for JUST how you simplified $\sqrt{76}$, not for the whole num'r.

You should have gotten: $\sqrt{76}=\sqrt{4\cdot 19}=\sqrt{4} \sqrt{19}=2\sqrt{19}$

Put that into the quadratic formula expression, and you get:

$\Large x=\dfrac{ -8 \pm 2\sqrt{19} }{ 2 }$

With me so far?

debbieg · Answer

Now you just need to reduce that, since you have a common factor of 2 in BOTH terms of the num'r, that cancels with the 2 in the den'r.

anonymous · Answer

okay so, $$x=-4\pm \sqrt{19}$$

debbieg · Answer

That's it!! Good job! :)

anonymous · Answer

oh my goodnesss, you're seriously the best! thank you!!!

debbieg · Answer

You were just off by that factor of 2 in front of the square root... and without seeing how you simplified the sq root, I couldn't determine if your error was in the simplification step, or in the reducing step. But as long as you understand it now.... it's all good! :)

debbieg · Answer

haha... you're welcome. Happy to help. :)