How do you find the discriminant for #2x^2-11x+10=0# and determine the number and type of solutions?

1 Answer
Jul 18, 2017

Answer:

#Delta = sqrt41#, so #f# has two real zeroes.

Explanation:

We know that, for any quadratic polynomial of the form #ax^2+bx+c#, the discriminant, #Delta#, is given by #Delta= b^2-4ac#.

If #∆ > 0#, then the polynomial will have two real solutions.

If #∆=0#, then the polynomial will have one real solution - a double root.

And if #∆ < 0#, then the polynomial will have no real roots and two complex roots.

And if #∆# is a perfect square, then the polynomial will have rational roots.

For the given polynomial, #∆=(11^2-4xx2xx10)=41#. Since #∆ > 0#, but #sqrt∆ notin QQ#, the polynomial will have two real roots.