using the quadriatic formula to solve!

4littlepiggiesmom · 2006-03-07 22:01:42

I have been working on this for a while and nothing I comes up with works if I try to put in to make sure it's right?

2x^2-x-2=0

I put them in the formula that looking like;

1 + or - the square root of 1+16 all over 4?

I've gottem 1.5, -1 4, 3.5 and some others.

What am I doing wrong???

RickyOswaldIOW · 2006-03-07 22:06:17

I think you're getting the quadratic formula wrong...

RickyOswaldIOW · 2006-03-07 22:13:04

x = -(b) +&- √(b)² - (4(a)(c)) / 2(a)

You get a, b and c from ax^2 +bx +c -> 2x^2 - 1x - 2
so a = 2, b = -1 and c = -2.

Put these into the quadratic formula:
-(-1) + √(-1)² - (4(2)(-2)) / 2(2) =
1 + √1² + 16 / 4 =
4.5

The working above shows you on solution for the +, you must change that to a - to work out the second solution!

Last edited by rickyoswaldiow (2006-03-07 22:32:00)

4littlepiggiesmom · 2006-03-07 22:21:04

Thanks I'm using that formula but correcting my sons homework and getting tired!

RickyOswaldIOW · 2006-03-07 22:41:54

Also, if you are using a calculator, make sure you are not doing 1 + √1² + (16/4), which would give you an incorrect answer. You can break the sum down:
√1² =
1
+ 16 =
17
+ 1 =
18
/4 =
4.5

RickyOswaldIOW · 2006-03-07 22:47:04

I think one of your errors is to do

the square root of 1+16

√(1+16) = √17, this is wrong. you should do:
(√1) + 16

4littlepiggiesmom · 2006-03-07 22:49:07

I've been using a calculator only if I really need to use it?? I see where your showing me though? I have 2 more problems on a nother post can you please check them for me??? Thanks again, Heidi

RickyOswaldIOW · 2006-03-07 22:54:38

I hate to say this now but I think I have made a mistake

RickyOswaldIOW · 2006-03-07 22:56:20

I think you are right to say

the square root of 1+16

My reasoning is incorrect, it would be pointless to find the square root of a number and then square it again (you end up with the same number)!

4littlepiggiesmom · 2006-03-07 22:58:39

Yeah I've tried that both ways but it seem easier for me?? Thanks agin! H

Jai Ganesh · 2006-03-07 23:29:10

Thje solutions according to the Quadratic Equation solver are x=1.2807764064044151 and x=-0.7807764064044151.

The Quadratic Equation solver is available on this website here.

Frank67625 · 2008-08-05 10:55:56

I've been working on this problem:

z(squared) = 9z - 1

The answer that I got was:

x = 9 +/- (square root of) 19.25

I'm not sure of my answer. Would you check my answer?
Thank you.

Sincerely,
Terry

TheDude · 2008-08-06 00:12:02

That answer is wrong Frank67625. You can easily check for yourself by plugging your answer into the equation. Using the positive square root you get z ~= 13.39. Substituting this value into your equation gives you

179.22 = 119.49, which is obviously wrong.

I'm not sure where you went wrong in coming up with that answer. Note that you can rewrite the equation to be z^2 - 9z + 1 = 0, which you can then use the quadratic formula on.

Also, please don't bump a 2 year old topic for a specific question like this. You should create a new thread to ask for help.

Math Is Fun Forum

#1 2006-03-07 22:01:42

using the quadriatic formula to solve!

#2 2006-03-07 22:06:17

Re: using the quadriatic formula to solve!

#3 2006-03-07 22:13:04

Re: using the quadriatic formula to solve!

#4 2006-03-07 22:21:04

Re: using the quadriatic formula to solve!

#5 2006-03-07 22:41:54

Re: using the quadriatic formula to solve!

#6 2006-03-07 22:47:04

Re: using the quadriatic formula to solve!

#7 2006-03-07 22:49:07

Re: using the quadriatic formula to solve!

#8 2006-03-07 22:54:38

Re: using the quadriatic formula to solve!

#9 2006-03-07 22:56:20

Re: using the quadriatic formula to solve!

#10 2006-03-07 22:58:39

Re: using the quadriatic formula to solve!

#11 2006-03-07 23:29:10

Re: using the quadriatic formula to solve!

#12 2008-08-05 10:55:56

Re: using the quadriatic formula to solve!

#13 2008-08-06 00:12:02

Re: using the quadriatic formula to solve!

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