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

2 Answers
Aug 20, 2015

Answer:

The solutions are
# color(blue)(x=-3#

# color(blue)(x=9#

Explanation:

#x^2−6x−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 =1*-27 = -27#
and
#N_1 +N_2 = b = -6#

After trying out a few numbers we get #N_1 = -9# and #N_2 =3#
#-9*3 = -27#, and #3+(-9)= -6#

#x^2−6x−27=x^2−9x+3x−27#

#x^2−9x+3x−27=0#

#x(x−9)+3(x−9)=0#
#(x+3) (x−9)=0# is the factorised form of the expression.

Now we equate these factors to the R.H.S (#0#)
#x+3=0, color(blue)(x=-3#

#x-9=0, color(blue)(x=9#

Aug 20, 2015

Answer:

Solve y = x^2 - 6x - 27 = 0

Ans: -3 and 9

Explanation:

x^2 - 6x - 27 = 0
Find 2 numbers knowing sum (6) and product (-27). Roots have opposite signs.
Factor pairs of -27 --> (-1, 27)(-3, 9). This last sum is 6 = -b. Therefor, the 2 real roots are: -3 and 9.