What are the number of real solutions to this equation: #1/3 x^2 - 5x +29 = 0#?

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Answer 1 · 2017-10-09T19:54:24

#0#

Given:

#1/3x^2-5x+29=0#

I'm not keen on doing more arithmetic than necessary with fractions. So let's multiply the whole equation by #3# to get:

#x^2-15x+87 = 0#

(which will have exactly the same roots)

This is in the standard form:

#ax^2+bx+c = 0#

with #a=1#, #b=-15# and #c=87#.

This has discriminant #Delta# given by the formula:

#Delta = b^2-4ac = (-15)^2-4(1)(87) = 225-348 = -123#

Since #Delta < 0# this quadratic equation has no Real roots. It has a Complex conjugate pair of non-real roots.