Question

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

Answer 1

`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`

Answer 2

`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"`