How to Write a Quadratic Equation Given Its Roots?

[abbarajum]

Hi everyone.

I was given a problem in which the roots of a quadratic function were given. Using those roots, I had to write the quadratic function, with integer coeffecients only.

The roots were: (-1+ (sqrt -2))/5 and its conjugate.

The equation I have so far is: f(x) = 5(x^2) + 2x + (3/5)

I used the sum and the product of roots to get this. My question is this. Is the constant term (3/5) considered a coeffecient? I ask this because I am not sure how to get all of these numbers to be integers. If it is considered a coeffecient, then should I use completing the square to rewrite the equation?

 

[SuperSonic4]

Hi everyone.

I was given a problem in which the roots of a quadratic function were given. Using those roots, I had to write the quadratic function, with integer coeffecients only.

The roots were: (-1+ (sqrt -2))/5 and its conjugate.

The equation I have so far is: f(x) = 5(x^2) + 2x + (3/5)

I used the sum and the product of roots to get this. My question is this. Is the constant term (3/5) considered a coeffecient? I ask this because I am not sure how to get all of these numbers to be integers. If it is considered a coeffecient, then should I use completing the square to rewrite the equation?

How did you get that equation? When I worked it out I got \(\displaystyle x^2 - \dfrac{2}{5}x + \dfrac{3}{25}\).

What you've done means the 5 is only distributed to the \(\displaystyle x^2\) instead the whole equation - a good way to check your work is to work out if the solution to the equation is what you started with

The constant term can be considered a coefficient (of \(\displaystyle x^0\)). Can you not multiply through by 5 again to give you: \(\displaystyle f(x) = \dfrac{1}{25}(25x^2-10x+3)\)