Question

Solve for x in the equation x^2-10x+25=35

Answer 1

x²-10x+25=35
(x-5)²=35
x-5=±√35
x=5±√35

Answer 2

A quadratic equation in the form of
`ax^2+bx+c = 0`....[1], then the solution of the equation is given by:

`x = (-b \pm √(b^2-4ac))/(2a)`

Given the equation:

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

Subtract 35 from both sides we have;

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

On comparing with [1] we have;

a = 1 , b = -10 and c = -10

then;

`x = (-(-10) \pm √((-10)^2-4(1)(-10)))/(2(1))`

⇒
`x = (10 \pm √(100+40))/(2)`

⇒
`x = (10 \pm √(140))/(2)`

Simplify:

`x = (10 \pm 2√(35))/(2)`

⇒
`x = 5 \pm √(35)`

Therefore, the value of x are:

`x = 5+ √(35)` and
`x = 5-√(35)`