SOLUTION: can i please have help to solve this? (x-1)^2=5

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Question 81000: can i please have help to solve this?
(x-1)^2=5
Found 2 solutions by praseenakos@yahoo.com, bucky:
Answer by praseenakos@yahoo.com(507) About Me  (Show Source):
You can put this solution on YOUR website!
QUESTION:

(x-1)^2=5


SOLUTION:


At first expand the expression using the identity (a-b)^2 = a^2 - 2ab + b^2 as follows............


(X-1)^2 = 5



==> x^2 - 2x + 1 = 5



subtract 5 from both sides




==> x^2 - 2x + 1 - 5 = 5 - 5



==> x^2 - 2x - 5 = 0



This is a quadratic equation so we can solve it using quadratic formula.



comparing with the standard equation, ax^2 + bx + c = 0, we have,


a = 1 b = -2 and c = -5


So solution is given by,



x+=+%28-%28-2%29+%2B-+sqrt%28+%28-2%29%5E2-4%2A1%2A%28-5%29+%29%29%2F%282%2A1%29+



x+=+%282+%2B-+sqrt%28+4%2B+20+%29%29%2F2+



x+=+%282+%2B-+sqrt%28+24%29%29%2F2+




x+=+%282+%2B-+4.9%29%2F2+



==> x = (2 + 4.9)/2 or x = (2-4.9)/2



==> x = 6.9/2 or x = -2.9/2



==> x = 3.45 or x = -1.45



To check your solution, substitute any of these values for x in the given equation, you will get tha answer 5, which is in the right hand side.



Hope you found this explanation useful.


Regards.


Praseena.



Answer by bucky(2189) About Me  (Show Source):
You can put this solution on YOUR website!
Given:
.
%28x-1%29%5E2=5
.
Square the left side and the equation becomes:
.
x%5E2+-+2x+%2B1+=+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%5E2+-+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%5E2+%2B+bx+%2B+c+=+0
.
The values of x that satisfy this equation are given by the equation:
.
x+=+%28-b%2B-sqrt%28b%5E2-4ac%29%29%2F%282a%29
.
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+=+%28-%28-2%29%2B-sqrt%28%28-2%29%5E2+-4%281%29%28-4%29%29%29%2F%282%2A1%29
.
This simplifies to:
.
x+=+%282%2B-sqrt%284+%2B+16%29%29%2F2+=+%282%2B-sqrt%2820%29%29%2F2+
.
Note that sqrt%2820%29+=+sqrt%284%2A5%29+=+sqrt%284%29%2Asqrt%285%29+=+2%2Asqrt%285%29
.
Substituting 2%2Asqrt%285%29 for sqrt%2820%29 reduces the equation for x to:
.
x+=+%282+%2B-+2%2Asqrt%285%29%29%2F2+=+1%2B-sqrt%285%29
.
So this problem has two solutions for x ... x+=+1+%2B+sqrt%285%29 and x+=+1+-+sqrt%285%29
.
Hope this helps you to understand the problem a little better.
.