How do you solve #10x^2 - 27x + 18#?

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Answer 1 · 2015-06-21T07:23:51

Use the quadratic formula to find zeros #x = 3/2# or #x=6/5#

#10x^2-27x+18 = (2x-3)(5x-6)#

#f(x) = 10x^2-27x+18# is of the form #ax^2+bx+c#, with #a=10#, #b=-27# and #x = 18#.

The discriminant #Delta# is given by the formula:

#Delta = b^2-4ac = 27^2 - (4xx10xx18) = 729 - 720#

#=9 = 3^2#

Being a positive perfect square, #f(x) = 0# has two distinct rational roots, given by the quadratic formula:

#x = (-b+-sqrt(Delta))/(2a) = (27+-3)/20#

That is:

#x = 30/20 = 3/2# and #x = 24/20 = 6/5#

Hence #f(x) = (2x-3)(5x-6)#

graph{10x^2-27x+18 [-0.25, 2.25, -0.28, 0.97]}

Answer 2 · 2015-06-21T07:24:35

#color(red)(x= 6/5 , x =3/2 #

#10x^2 - 27x + 18 = 0#

We can first factorise the above expression and thereby find the solution.

Factorising by splitting the middle term

#10x^2 - 15x - 12x + 18 = 0#
# 5x(2x- 3) - 6(2x - 3) = 0#
#color(red)( (5x - 6)(2x- 3) ) = 0#

Equating each of the two terms with zero we obtain solutions as follows:

#color(red)(x= 6/5 , x =3/2 #