Question

Last question ..

Answer 1

@hartnn

Answer 2

lets take x^2+x-1=0(cannot be solved by factoring)
Compare your quadratic equation with \(ax^2+bx+c=0\)
find a,b,c
then the two roots of x are:
\(\huge{x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}}\)

Answer 3

To have a root b^2-4ac>=0
And for the quadratic equation not to solved by factoring b^2-4ac should not be a perfect square number.

Answer 4

x^2+x-1=0(cannot be solved by factoring) because there are no pair of numbers with product =-1 and sum = 1

Answer 5

@hartnn , what could I put as the a,b, and c value ?

Answer 6

x^2+x-1=0
a-1 , b=1 , c=-1

Answer 7

Thanks