# Thread: Quadratic Equation question

1. ## Quadratic Equation question

Hello, I wasn't sure if this question belonged here or under another section. In the grade 12 math text I'm using, I've learned pretty much 3 main ways to solve quadratic equations: 1) Synthetic Division, 2) the Factor Theorem, and 3) the Quadratic Formula. The problem I have I will show in a few steps:

The questions:

1) 3t^3 - t^2 - 6t + 2 = 0

The roots are: 1/3; and + or - (square root of 2)

2) 2m^3 - 5m^2 + 1 = 0

The roots are: 1/2; and 1 + or - (square root of 2)

The problem:

So far, the main approach I use is the factor theorem for quadratics with higher than a power of 2, so I can get the roots and determine the factors. For example:

x^4 - 27x = 0
R=P(3)=(3)^4 - 27(3)
=(3)27-(3)27=0
Therefore, 3 is a root, meaning (x-3) is a factor.

The problem I have is using this process I just showed you for the examples 1) and 2) above. Basically, I use factors of the # with the smallest power of x: i.e. 27x, I would test 1,-1, 3, -3, 9, -9, 27, -27.

The problem is that I don't have a good method of determining the roots for 1) and 2), since they are fractions and/or square roots, and since I have to substitute them into x^2 and x^3, it's a bit difficult for me to see the answer right away. Is there a fairly simple method to finding these that anyone could show me? Strangely enough, out of over 200 questions this chapter, these are the only two like this, and I can't solve them because I can't find a factor to work with in order to continue to the next steps.

Thanks for your time, sorry about the lengthy post.

W

Edit: oops, minor mistake

2. Originally Posted by WRob99
Hello, I wasn't sure if this question belonged here or under another section. In the grade 12 math text I'm using, I've learned pretty much 3 main ways to solve quadratic equations: 1) Synthetic Division, 2) the Factor Theorem, and 3) the Quadratic Formula. The problem I have I will show in a few steps:

The questions:

1) 3t^3 - t^2 - 6t + 2 = 0

The roots are: 1/3; and + or - (square root of 2)
The rational root theorem tells you that a rational root of this is the ratio of a factor of the constant term to that of a factor of the coefficient of the highest power of the variable.

So the candidates are 2, 1, 2/3, 1/3, try them in turn, if one is a root then take out the corresponding factor leaving you with a quadratic. Now use the quadratic formula, complete the square or whatever to find the roots of this quadratic.

CB

3. Thanks CB, that's very helpful. Strange how it wasn't in the same chapter, although I suspect that it might have been edited out or something, since the text seems to have been shortened.

W