Asked by Lucy
2+3X^2-X^3=0 by use quadratic formula. The answer of textbook is 3.19
So I solved and came out a wrong answer
What I did factor out
(-X+3)X^2+2=0
Then By quadratic formula, a=1,b=0,c=2. Thus I got sqrt(8)/2
So I solved and came out a wrong answer
What I did factor out
(-X+3)X^2+2=0
Then By quadratic formula, a=1,b=0,c=2. Thus I got sqrt(8)/2
Answers
Answered by
Reiny
you can't use the quadratic formula on
2+3X^2-X^3=0 , since it is not a quadratic, it is a cubic
Do you have a typo?
Assuming it is correct, lets see if it factors:
let f(x) = 2 + 3x^2 - x^3
f(1) = 2 + 3 - 1 ? 0
f(-1) = 2 + 3 - (-1) ?0
f(2) = 2 + 12 - 8 ? 0
f(-2) = 2 + 12 + 8 ? 0
no rational factors
let's see what Wolfram says:
http://www.wolframalpha.com/input/?i=solve+2%2B3x%5E2+-+x%5E3++%3D+0
click on approximate forms to get x = 3.1958
and two complex roots.
There are several methods to solve a cubic, but the quadratic formula is certainly not one of them.
What you did is bogus.
2+3X^2-X^3=0 , since it is not a quadratic, it is a cubic
Do you have a typo?
Assuming it is correct, lets see if it factors:
let f(x) = 2 + 3x^2 - x^3
f(1) = 2 + 3 - 1 ? 0
f(-1) = 2 + 3 - (-1) ?0
f(2) = 2 + 12 - 8 ? 0
f(-2) = 2 + 12 + 8 ? 0
no rational factors
let's see what Wolfram says:
http://www.wolframalpha.com/input/?i=solve+2%2B3x%5E2+-+x%5E3++%3D+0
click on approximate forms to get x = 3.1958
and two complex roots.
There are several methods to solve a cubic, but the quadratic formula is certainly not one of them.
What you did is bogus.
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