For the equation ax^2+bx+c=0ax2+bx+c=0, the roots are given by x=(-b+-sqrt(b^2-4ac))/(2a)x=−b±√b2−4ac2a. As can be seen type of roots depend upon discriminant b^2-4acb2−4ac. If is negative, roots are complex and if it is a complete square (and coefficients are rational), roots are rational.
In the equation 4x^2+4x-8=14x2+4x−8=1 or 4x^2+4x-9=04x2+4x−9=0, the discriminant is (4)^2-4×4×(-9)=16+144=160(4)2−4×4×(−9)=16+144=160, which is not a complete square and hence roots are irrational.
Roots are x=(-4+-sqrt160)/(2×4)x=−4±√1602×4
= (-4+-4sqrt10)/8−4±4√108
= (-1+-sqrt10)/2−1±√102
Hence solution is (-1-sqrt10)/2−1−√102 and (-1+sqrt10)/2−1+√102.
