Question

How do you solve `(5x-1)^2=16`?

Answer 1

`x=-3/5` and `x=1`

Explanation:

First you expand `(5x-1)^2` so `(5x-1)*(5x-1) = 25x^2-10x +1`
That leaves us with:
`25x^2-10x +1=16`
Now we subtract 16 from both sides:
`25x^2-10x -15=0`
Then all we have to do is plug a,b and c from `ax^2 +bx+c` into the quadratic formula `x=(-b+-sqrt(b^2-4ac))/(2a)`
`a=25`
`b=-10`
`c=-15`

`x=(-(-10)+-sqrt((-10)^2-4*25*(-15)))/(2*25)`

`x=(10+-sqrt(100-(-1500)))/50`

`x=(10+-sqrt(1600))/50`

`x=(10+-40)/50`

So `x=50/50=1` and `x=-30/50=-3/5`

Answer 2

Take the square root of both sides and solve to get to `x=1`

Explanation:

Start with:

`(5x-1)^2=16`

The first problem we have is that everything on the left is squared. We could multiply it out and end up with `25x^2-10x+1`, but since the right side is a perfect square, let's instead take the square root of both sides:

`sqrt((5x-1)^2)=sqrt16`

`5x-1=4`

And now let's solve:

`5x=5`

`x=1`

And we can check the answer:

`(5x-1)^2=16`

`(5(1)-1)^2=16`

`(5-1)^2=16`

`(4)^2=16`

`16=16`