How do you solve #6x^2 - 27x +27= 0# by factoring?

Question

3003 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-08-20T13:13:44

The solutions are
#color(blue)(x=3/2#

# color(blue)(x=3#

#6x^2−27x+27=0#

We can Split the Middle Term of this expression to factorise it and thereby find solutions.

In this technique, if we have to factorise an expression like #ax^2 + bx + c#, we need to think of 2 numbers such that:

#N_1*N_2 = a*c = 6*27 = 162#
and
#N_1 +N_2 = b = -27#

After trying out a few numbers we get #N_1 = -18# and #N_2 =-9#
#-18*-9 = 162#, and #(-18)+(-9)=-27#

#6x^2−27x+27=6x^2−18x-9x+27#
#6x(x-3) -9(x-3) =0#

#(6x-9)(x-3) =0# is the factorised form of the expression.

We now equate each of these factors to the R.H.S (#0#)
#6x-9=0, x=9/6,color(blue)(x=3/2#
#x-3=0, color(blue)(x=3#