How do you find the discriminant and how many and what type of solutions does #4x^2-7x+2=0# have?

1 Answer
Mar 19, 2018

Answer:

We have two conjugate irrational numbers of type #a+-sqrtb#.

Explanation:

The discriminant of a quadratiic equation #ax^2+bx+c=0# is #b^2-4ac#. Assume that coefficients #a#, #b# and #c# are all integers, then

  1. If #b^2-4ac# is equal to zero, we have just one solution
  2. If #b^2-4ac>0# and it is square of a rational number, we have two solutions, each a rational number.
  3. If #b^2-4ac>0# and it is not a square of a rational number, we have two solutions, each root being a real number but not rational number. Tey will be of type #a+-sqrtb#, two conjugate irrational numbers.
  4. If #b^2-4ac<0#, we have two solutions, which are two complex conjugate numbers.

Here in #4x^2-7x+2-0#, we have #a=4,b=-7# and #c=2# and hence dicriminant is #(-7)^2-4*4*2=49-32=17#

we have two conjugate irrational numbers of type #a+-sqrtb#