How do you find the nature of the roots using the discriminant given #x^2 - 7x + 12 = 0#?

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Answer 1 · 2016-09-25T17:21:45

Roots are rational. These are #x=3# or #x=4#.

If the equation is #ax^2+bx+c=0#, nature of roots is decided by discriminant #b^2-4ac#

If #a#, #b# and #c# are rational and #b^2-4ac# is square of a rational number, roots are rational.

If #b^2-4ac>0# but is not a square of a rational number, roots are real but not rational.

If #b^2-4ac>0-0# we have equal roots.

If #b^2-4ac<0# roots are complex

In #x^2-7x+12=0#, discriminant is #(-7)^2-4xx1xx12=49-48=1=1^2#

hence roots are rational. In fact

#x^2-7x+12=0#

#hArrx^2-4x-3x+12=0#

or #x(x-4)-3(x-4)=0#

or #(x-3)(x-4)=0# i.e. #x=3# or #x=4#