What kind of solutions does #7R2 -14R + 10 = 0# have?

1 Answer
Jul 5, 2015

Answer:

#7R^2-14R+10# has discriminant #Delta = -84 < 0#.

So #7R^2-14R+10=0# has no real solutions.

It has two distinct complex solutions.

Explanation:

#7R^2-14R+10# is of the form #aR^2+bR+c# with #a=7#, #b=-14# and #c=10#.

This has discriminant #Delta# given by the formula:

#Delta = b^2-4ac = (-14)^2-(4xx7xx10) = 196 - 280 = -84#

Since #Delta < 0# the equation #7R^2-14R+10 = 0# has no real roots. It has a pair of complex roots that are complex conjugates of one another.

The possible cases are:
#Delta > 0# The quadratic equation has two distinct real roots. If #Delta# is a perfect square (and the coefficients of the quadratic are rational), then those roots are also rational.

#Delta = 0# The quadratic equation has one repeated real root.

#Delta < 0# The quadratic equation has no real roots. It has a pair of distinct complex roots that are complex conjugates of one another.