SOLUTION: Given that the quadratic equation x^2-3x+1=0 has 2 different roots, create a quadratic equation, so that each of its roots will be 1 more than the corresponding root of the given q

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: Given that the quadratic equation x^2-3x+1=0 has 2 different roots, create a quadratic equation, so that each of its roots will be 1 more than the corresponding root of the given q      Log On


   

Question 839474: Given that the quadratic equation x^2-3x+1=0 has 2 different roots, create a quadratic equation, so that each of its roots will be 1 more than the corresponding root of the given quadratic equation.
Answer by josh_jordan(263) About Me  (Show Source):
You can put this solution on YOUR website!
The first thing we need to do is find the roots of x%5E2-3x%2B1=0.

You will notice that this polynomial will not factor evenly, so we will have to find the roots by using the quadratic formula: x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+

Doing this will give us our two roots: %28%283%2Bsqrt%285%29%29%2F2%29 and %28%283-sqrt%285%29%29%2F2%29

To find a new quadratic equation with roots that are one more than each of the roots of our original equation, we need to add 1 to each of the original roots:

%28%283%2Bsqrt%285%29%29%2F2%29%2B1 = %28%285%2Bsqrt%285%29%29%2F2%29 and

%28%283-sqrt%285%29%29%2F2%29%2B1 = %28%285-sqrt%285%29%29%2F2%29

Next, we will put these roots into factored form:

%28x-%28%285%2Bsqrt%285%29%29%2F2%29%29%2A%28x-%28%285-sqrt%285%29%29%2F2%29%29

Finally, we need to expand our factors by multiplying via the FOIL method, which will give us our final answer:

x%5E2-5x%2B5=0