Question 81000
Given:
.
{{{(x-1)^2=5}}}
.
Square the left side and the equation becomes:
.
{{{x^2 - 2x +1 = 5}}}
.
Then get this equation into standard quadratic form by eliminating the 5 on the right side
so that the right side becomes 0.  Do this by subtracting 5 from both sides.  When you do
that subtraction the equation becomes:
.
{{{x^2 - 2x - 4 = 0}}}
.
The left side of this equation does not factor nicely. So use the quadratic formula. This
formula says that for a quadratic equation of the standard form:
.
{{{ax^2 + bx + c = 0}}}
.
The values of x that satisfy this equation are given by the equation:
.
{{{x = (-b+-sqrt(b^2-4ac))/(2a)}}}
.
By comparing our equation with the standard form you can see that a = 1, b = -2, and c = -4.
.
By substituting these values into the equation that defines the values of x you get:
.
{{{x = (-(-2)+-sqrt((-2)^2 -4(1)(-4)))/(2*1)}}}
.
This simplifies to:
.
{{{x = (2+-sqrt(4 + 16))/2 = (2+-sqrt(20))/2 }}}
.
Note that {{{sqrt(20) = sqrt(4*5) = sqrt(4)*sqrt(5) = 2*sqrt(5)}}}
.
Substituting {{{2*sqrt(5)}}} for {{{sqrt(20)}}} reduces the equation for x to:
.
{{{x = (2 +- 2*sqrt(5))/2 = 1+-sqrt(5)}}}
.
So this problem has two solutions for x ... {{{x = 1 + sqrt(5)}}} and {{{x = 1 - sqrt(5)}}}
.
Hope this helps you to understand the problem a little better.
.