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"