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

Mar 19, 2018

We have two conjugate irrational numbers of type $a \pm \sqrt{b}$.

#### Explanation:

The discriminant of a quadratiic equation $a {x}^{2} + b x + c = 0$ is ${b}^{2} - 4 a c$. Assume that coefficients $a$, $b$ and $c$ are all integers, then

1. If ${b}^{2} - 4 a c$ is equal to zero, we have just one solution
2. If ${b}^{2} - 4 a c > 0$ and it is square of a rational number, we have two solutions, each a rational number.
3. If ${b}^{2} - 4 a c > 0$ 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 $a \pm \sqrt{b}$, two conjugate irrational numbers.
4. If ${b}^{2} - 4 a c < 0$, we have two solutions, which are two complex conjugate numbers.

Here in $4 {x}^{2} - 7 x + 2 - 0$, we have $a = 4 , b = - 7$ and $c = 2$ and hence dicriminant is ${\left(- 7\right)}^{2} - 4 \cdot 4 \cdot 2 = 49 - 32 = 17$

we have two conjugate irrational numbers of type $a \pm \sqrt{b}$