Question

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

Answer 1

The left side of this equation is already a perfect square: x^2-10x+25=35.

Rewriting it, we get (x-5)^2 = 35.

Taking the sqrt of both sides, x-5 = sqrt(35).

Solving for x: x = 5 plus or minus sqrt(35) (answers)

Answer 2

`\text{The solution is }x=5\pm √(35)`

Explanation:

Given the equation

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

we have to solve the above equation for x

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

Subtracting 35 from both sides, we get

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

`\text{Comparing above equation with }ax^2+bx+c=0`

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

By quadratic formula, the solution is

`x=(-b\pm √(b^2-4ac))/(2a)`

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

`x=(10\pm √(100+40))/(2)`

`x=(10 \pm √(140))/(2)`

`x=(10 \pm 2√(35))/(2)`

`x=5\pm √(35)`

which is required solution.