How do you find the real solutions of the polynomial #2x^3=19x^2-49x+20#?

1 Answer
Feb 20, 2017

Answer:

The roots are #1/2#, #5# and #4#.

Explanation:

Given:

#2x^3=19x^2-49x+20#

Subtract the right hand side from the left to get the standard form:

#2x^3-19x^2+49x-20 = 0#

By the rational root theorem, any rational zeros of this cubic are expressible in the form #p/q# for integers #p, q# with #p# a divisor of the constant term #-20# and #q# a divisor of the coefficient #2# of the leading term.

That means that the only possible rational roots are:

#+-1/2, +-1, +-2, +-5/2, +-4, +-5, +-10, +-20#

We find that #x = 1/2# is a zero...

#2(color(blue)(1/2))^3-19(color(blue)(1/2))^2-49(color(blue)(1/2))+20 = (1-19-98+80)/4 = 0#

Hence #(2x-1)# is a factor:

#2x^3-19x^2+49x-20 = (2x-1)(x^2-9x+20)#

To factor the remaining quadratic, find a pair of factors of #20# with sum #9#. The pair #5, 4# works and hence:

#x^2-9x+20 = (x-5)(x-4)#

So the other two roots are #x=5# and #x=4#.