How do you solve #\x ^ { 2} - 8x = 20#?

2 Answers
Jun 2, 2017

#x = -2 or x= 10#

Explanation:

#x^2 - 8x = 20#

Re-arrange to form a standard quadratic form
#x^2 - 8x - 20 = 0#

Find the factors of #-20# - such as #(-10 xx2)#

where sum of factors = #-8# again they are #(2 + (-10))#

and rearrange in the equation.
#x^2 + 2x - 10x - 20 = 0 #

#x(x +2) -10(x + 2) =0#

#(x + 2)(x - 10) = 0#

Setting each factor equal to #0#

#x = -2 or x= 10#

Jun 2, 2017

#x=-2" or " x=10#

Explanation:

#"rearrange and equate to zero"#

#rArrx^2-8x-20=0#

#"factorise by 'splitting' the middle term"#

#rArrx^2+2x-10x-20=0larr" 2x - 10x = -8x"#

#"take out a common factor from each 'pair' of terms"#

#rArrcolor(red)(x)(x+2)color(red)(-10)(x+2)=0#

#"take out the common factor of " (x+2)#

#rArr(x+2)(color(red)(x-10))=0#

#"equate each factor to zero and solve"#

#x+2=0rArrx=-2larr" solution"#

#x-10=0rArrx=10larr" solution"#