How do you factor and solve #x^2+2x-8=0#?

2 Answers
Don't Memorise
Jun 10, 2016

The solutions for the equation are #color(blue)(x=2 , color(blue)(x=-4#

Explanation:

#x^2 + 2x - 8 = 0#

We can Split the Middle Term of this expression to factorise it.

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*(-8) = -8#

AND

#N_1 +N_2 = b = 2#

After trying out a few numbers we get #N_1 = -2# and #N_2 =4#
#(-2)*4 = -8#, and #-2+ 4= 2#

#x^2 + color(blue)( 2x) - 8 = x^2 + color(blue)(4x-2x) - 8 #

# = x(x+4) - 2(x+4)#

#color(blue)((x+4))# is a common factor to each of the terms

# color(green)(( x - 2 )(x+4) # is the factorised form of the expression. Now we equate the factors to zero.

  • #color(blue)(x-2) =0 , color(blue)(x=2 #

  • #color(blue)(x + 4) =0 , color(blue)(x=-4#

Nghi N.
Jun 10, 2016

2 and -4

Explanation:

Solve #y = x^2 + 2x - 8.#
The 2 real roots have opposite signs because ac < 0
Find 2 real roots knowing sum (-b = -2) and product (c = -8).
Factor pairs of (c = -8) --> (-2, 4)(2, -4). This last sum is (-2 = -b).
Therefor, the 2 real roots are: 2 and -4.

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