Question

How do you find the discriminant, describe the number and type of root, and find the exact solution using the quadratic formula given `2x-5=-x^2`?

Answer 1

Roots are `-1+sqrt6` and `-1-sqrt6`

Explanation:

For an equation `ax^2+bx+c=0`, discriminant is `b^2-4ac`

and quadratic formula gives roots as `x=(-b+-sqrt(b^2-4ac))/(2a)`

Note that if `a,b`and `c` are rational number, as we have in the given example

if `b^2-4ac` is a perfect square (i.e. positive as well), roots are rational

if `b^2-4ac` is positive but not a perfect square, roots are irrational and conjugate i.e. of the type `p+-sqrtq`

if `b^2-4ac` is negative, roots are complex conjugate i.e. of the type `p+-iq`

Here we have `2x-5=-x^2` i.e. `x^2+2x-5=0`

and as `a=1`, `b=2` and `c=-5`, the discriminant is

`b^2-4ac=2^2-4xx1xx(-5)=4+20=24`

and hence roots are irrational and conjugate.

These are `(-2+-sqrt24)/2=-2/2+-(2sqrt6)/2=-1+-sqrt6`