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

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Answer 1 · 2018-03-19T08:55:41

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

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

If $b^2-4ac$ is equal to zero, we have just one solution
If $b^2-4ac>0$ and it is square of a rational number, we have two solutions, each a rational number.
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.
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 $a+-sqrtb$

How do you find the discriminant and how many and what type of solutions does 4x^2-7x+2=04x^2-7x+2=0 have?