Using the Quadratic Formula (Real Solution)

[holmes1172]

Checking my work:

x²-3-2=0

[FONT=&quot][/FONT][FONT=&quot][/FONT] x^2-3x-2=0
x=-b±√(b^2-4ac)/2a
X=-(-3)±√(-3)^2+4(1)(-2)/2(1)
X=3±√(9-8)/2
X=3√9-4

I feel I should simplify once more and take out three dividing the 9 down to 3 to have something like 3-4...am I at least on the right track with most of my work? Sure appreciate this forum's help.

 

[Quaid]

Checking my work:

x² - 3 - 2 = 0

x^2 - 3x - 2 = 0

x = -b ± √(b^2 - 4ac)/(2a)

X = -(-3) ± √[(-3)^2 + 4(1)(-2)]/[2(1)]

X = 3 ± √(9 - 8)/2

X = 3√9 - 4

Typographical mistakes shown in green, and algebra errors shown in red.

Your error began when you wrote + instead of - in line 4.

Fix that, try again, and show us what you get. (You do not need to retype the previous lines.)

take out three dividing the 9 down to 3 to have something like 3-4

What language is that, lol?

am I at least on the right track with most of my work?

Mostly, yes.

I am concerned that you seemed to simplify 8/2 to 4, in the last line. You may not do that. Follow the Order of Operations (i.e., complete the arithmetic inside the radical sign BEFORE dividing by 2).

Please proofread your typing, before submitting your posts; use the [Preview Post] button for that.

Grouping symbols are required, as shown above.

Do not use x and X to represent the same variable; stick with one or the other.

Cheers :cool:

 

[holmes1172]

X= - (-3) ± √ [(-3)^2 -4 (1) (-2)] / [2 (1)]

x= 3 ± √ (9+8) / 2

x= 3 √ 15 / 2

Am I on the right track now? Hopefully I was able to clean this up a bit.

 

[Quaid]

X = -(-3) ± √[(-3)^2 -4(1)(-2)] / [2(1)]

x= 3 ± √(9+8) / 2

x= 3 ? √15 / 2

Am I on the right track now?

You're moving forward, but you're still off track.

What is 9+8?

What replaces the ? above? (It's not multiplication.)

:cool:

Hopefully I was able to clean this up a bit.

Your formatting is getting much better. Thank you for that!

 

[holmes1172]

You're moving forward, but you're still off track.

What is 9+8?

What replaces the ? above? (It's not multiplication.)

:cool:




Your formatting is getting much better. Thank you for that!

DOH!! 17. Simple mistakes like that that really cost me the right answers because I wasn't paying attention to detail. Thanks for that catch. Too much algebra today. lol

X = 3 ± √ (9+8) / 2


X = 3 ± √ 17?

But then where do I need to go from here? And was it the ± I needed to replace the "?" ?

 

[JeffM]

X= - (-3) ± √ [(-3)^2 -4 (1) (-2)] / [2 (1)] You still are not grasping the importance of order of operations.

- (-3) ± √ [(-3)^2 -4 (1) (-2)] / [2 (1)] = \(\displaystyle - (-3) \pm \dfrac{\sqrt{(-3)^2 - 4(1)(-2)}}{2 * 1}.\) That is not what you meant.

{- (-3) ± √ [(-3)^2 -4 (1) (-2)]} / [2 (1)] = \(\displaystyle \dfrac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(-2)}}{2 * 1}.\). That is what you meant.

x= 3 ± √ (9+8) / 2

x= 3 √ 15 / 2 \(\displaystyle 9 + 8 \ne 15.\) Be careful. Your algebra can be perfect, but your arithmetic has to be correct too.

Furthermore, what happened to the \(\displaystyle \pm\).

Am I on the right track now? Hopefully I was able to clean this up a bit.

I think you are in the neighborhood of the right track. Unfortunately, math requires careful attention to detail. We ALL make mistakes, but you must try to minimize them. And then you must check your work.

Do you know how to check your answers to a quadratic? Why don't you try again, and then show US how your answers check. That way, you and we will ALL be sure that you know how to do the work and check it.

 

[Quaid]

X = 3 ± √ (9+8) / 2

X = 3 ± √ 17

What happened to the division by 2, on the last line?

was it the ± I needed to replace the "?" ?

Yes.

Writing 3√17 means: 3 times √17

That's not what the Quadratic Formula says...

 

[holmes1172]

I think you are in the neighborhood of the right track. Unfortunately, math requires careful attention to detail. We ALL make mistakes, but you must try to minimize them. And then you must check your work.

Do you know how to check your answers to a quadratic? Why don't you try again, and then show US how your answers check. That way, you and we will ALL be sure that you know how to do the work and check it.

How would I check my answer on something like this? Usually, when I check my answer, it's a certain value I can just plug in, but here I'm not sure.

 

[holmes1172]

What happened to the division by 2, on the last line?




Yes.

Writing 3√17 means: 3 times √17

That's not what the Quadratic Formula says...

Oh jeeze. I'm losing it. Dopey mistakes I'm making.

X = 3 ± √ 17 / 2

Would that be the final answer then? Or do I need to go ahead and divide 17 by 2 to get as the final step?

 

[Quaid]

X = [3 ± √17]/2

Would that be the final answer then? Or do I need to go ahead and divide 17 by 2 to get as the final step?

That's the final answer.

17 is not being divided by 2 here.

It's the square root of 17 that is being divided by 2.

\(\displaystyle \dfrac{\sqrt{17}}{2} = \dfrac{4.1231}{2}\)

We would only do this if we were asked to report decimal approximations.

---

The answer can also be written as:

\(\displaystyle x = \dfrac{3}{2} \pm \dfrac{\sqrt{17}}{2}\)

 

[Quaid]

I'm losing it. Dopey mistakes I'm making.

I'm going for a walk; I think that you should, too. :cool:

 

[holmes1172]

That's the final answer.

17 is not being divided by 2 here.

It's the square root of 17 that is being divided by 2.


\(\displaystyle \dfrac{\sqrt{17}}{2} = \dfrac{4.1231}{2}\)

We would only do this calculation if we were asked to report decimal approximations.

No, they didn't. So, in the future, unless they DO ask, we would normally leave it right there? What if it was something evenly divisible we were left with like 16 / 8? Would we go ahead divide it down to get a single number?

 

[Quaid]

You should always reduce fractions to lowest terms, when reporting an answer.

So, yes. 16/8 would be written as 2.

The point I'm trying to make is that you must follow the Order of Operations.

If you were told to report a decimal value, then you must take the square root of 17 BEFORE doing the division.

 

[JeffM]

How would I check my answer on something like this? Usually, when I check my answer, it's a certain value I can just plug in, but here I'm not sure.

For some reason, it takes a while to understand that radicals are just numbers.

Here is how to check the answers to a quadratic

\(\displaystyle x^2 - 5x - 14 = 0 \implies \dfrac{(-5) \pm \sqrt{(-5)^2 - 4(1)(-14)}}{2 * 1} =\dfrac{5 \pm \sqrt{25 + 56}}{2} = \dfrac{5 \pm \sqrt{81}}{2} = \dfrac{5 + 9}{2}\ or\ \dfrac{5 - 9}{2} = 7/ or/ - 2.\)

You check by putting your answers back into the original equation.

\(\displaystyle 7^2 - 5 * 7 - 14 = 49 - 35 - 14 = 14 - 14 = 0.\) That answer checks.

\(\displaystyle (-2)^2 - 5(-2) - 14 = 4 + 10 - 14 = 14 - 14 = 0.\) That answer checks too.

Got that idea? You do the same thing if the answer contains a radical.

Now if you get an answer with a radical , it's actually easier because you only have to check one of the two answers.

\(\displaystyle x^2 + 3x + 1 = 0 \implies x = \dfrac{- 3 \pm \sqrt{3^2 - 4 * 1 * 1}}{2 * 1} = \dfrac{-3 \pm \sqrt{9 - 4}}{2} = \dfrac{- 3 \pm \sqrt{5}}{2}.\)

It turns out when we have symmetric answers like this, we need to check only one answer to confirm both answers.

\(\displaystyle \left(\dfrac{- 3 + \sqrt{5}}{2}\right)^2 + 3\left(\dfrac{- 3 + \sqrt{5}}{2}\right) + 1 = \dfrac{9 + 2(- 3) \sqrt{5} + 5}{4} + \dfrac{- 9 + 3\sqrt{5}}{2} + 1 = \dfrac{9 - 6\sqrt{5} + 5}{4} + \dfrac{- 18 + 6\sqrt{5}}{4} +\dfrac{4}{4} = \dfrac{9 + 5 + 4 - 18}{4} = 0.\)

That answer checks. Fussing with the radicals makes the arithmetic a bit harder so it is a blessing that you only have to check one of the two answers if they contain radicals.