Question

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

Answer 1

Answer:

$\text{The solution is }x=5\pm \sqrt{35}$    

Step-by-step 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

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$x=\frac{-(-10)\pm \sqrt{(-10)^2-4(1)(-10)}}{2(1)}$

$x=\frac{10\pm \sqrt{100+40}}{2}$

$x=\frac{10 \pm \sqrt{140}}{2}$

$x=\frac{10 \pm 2\sqrt{35}}{2}$

$x=5\pm \sqrt{35}$

which is required solution.

Answer 2

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)