Question

How do you solve `x^2-10x+25=35`?

Answer 1

`x = 5 + sqrt(35)`
or
`x = 5 - sqrt(35)`

Explanation:

1. Take 35 from both sides so that the quadratic equation is equal to 0.
   `x^2 -10x - 10 = 0`

2. Solve for x using the following equation:
   `x = (-b +- sqrt(b^2 -4ac))/(2a)`
   `= (-(-10) +- sqrt((-10)^2 -4(1)(-10)))/(2(1))`
   `= (10 +- sqrt(100 +40))/2`
   `= (10 +- sqrt(140))/2`
   `= (10 +- (sqrt(35)*sqrt(4)))/2`
   `= (2(5) +- 2(sqrt(35)))/2`
   `= (2(5 +- sqrt(35)))/2`
   `= 5 +- sqrt(35)`

Answer 2

`x=5+-sqrt35`

Explanation:

`color(blue)(x^2-10x+25=35`

`rarrx^2-10x+25-35=0`

`rarrx^2-10x-10=0`

This is a quadratic equation (in form of `color(orange)(ax^2+bx+c=0`)

We use the quadratic formula to solve it

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

Where,

`color(violet)(a=1`

`color(violet)(b=-10`

`color(violet)(c=-10`

`rarrx=(-(-10)+-sqrt((-10)^2-4(1)(-10)))/(2(1))`

`rarrx=(10+-sqrt(100-(-40)))/(2)`

`rarrx=(10+-sqrt(100+40))/(2)`

`rarrx=(10+-sqrt(140))/(2)`

`rarrx=(10+-sqrt(4*35))/(2)`

`rarrx=(10+-2sqrt(35))/(2)`

`rarrx=(cancel10^5+-cancel2^1sqrt(35))/(cancel2^1)`

`rArrcolor(green)(x=5+-sqrt35`

Answer 3

`x = 5+-sqrt(35)`

Explanation:

Notice that the left hand side of the equation is already a perfect square trinomial.

So rather than reformulate, we can proceed as follows:

`(x-5)^2 = x^2-10x+25 = 35`

Take the square root of both ends, allowing for both positive and negative square roots to get:

`x-5 = +-sqrt(35)`

Add `5` to both sides to find:

`x = 5+-sqrt(35)`

Finally note that `35=5*7` has no square factors, so `sqrt(35)` is already in simplest form.