Question

How do you solve `x^2 - 4x = 5` using the quadratic formula?

Answer 1

`color(blue)(x=-1 or x=5`

Explanation:

General quadratic equation:

`ax^2+bx+c=0`

Quadratic formula:

`x=(-b+-sqrt(b^2-4ac))/(2a)`

`x^2-4x=5`

`:.x^2-4x-5=0`

`:.a=1,b=-4,c=-5`

`:.x=(-(-4)+-sqrt((-4)^2-4(1)(-5)))/(2(1))`

`:.x=(4+-sqrt(16+20))/2`

`:.x=(4+-sqrt36)/2`

`:.x=(4+-6)/2`

`:.x=(4+6)/2`

or:

`:.x=(4-6)/2`

`:.x=10/2` or `x=(-2)/2`

`:.color(blue)(x=5` or `color(blue)(x=-1`

substitute `color(blue)(x=-1`

`:.(-1)^2-4(-1)=5`

`:.1+4=5`

`:.color(blue)(color(blue)(5=5`

substitute `color(blue)(x=5`

`:.(5)^2-4(5)=5`

`:.25-20=5`

`:.color(blue)(5=5`

Answer 2

`x=5, x=-1`

Explanation:

Subtract `5` from both sides.

`x^2-4x-5=0`

From this, `a=1, b=-4, c=-5`

The quadratic formula is `(-b+-sqrt(b^2-4ac))/(2a)`

Substituting the values

`(-(-4)+-sqrt((-4)^2-4*1*-5))/(2*1)`

`(4+-sqrt(16+20))/2`

`(4+sqrt(36))/2`

`4/2+sqrt(36)/2, 4/2-sqrt(36)/2`

`2+sqrt(36)/2, 2-sqrt(36)/2`

`2+6/2, 2-6/2`

`2+3, 2-3`

`5, -1`