SOLUTION: use a difference of squares method to solve the following: {{{x^2-5x+3=x(x-5)+3=0}}} then substitute A=x-2.5

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: use a difference of squares method to solve the following: {{{x^2-5x+3=x(x-5)+3=0}}} then substitute A=x-2.5      Log On


   



Question 1086778: use a difference of squares method to solve the following:
x%5E2-5x%2B3=x%28x-5%29%2B3=0 then substitute A=x-2.5

Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
this dosn't look like a difference of square method.

different of square is (a^2 - b^2) = (a-b) * (a+b)

i don't see that in this problem.

your equation is x^2 - 5x + 3 = 0.

the roots are not integers.

you can solve this graphically or by use of the quadratic formula, but not by difference of squares as far as i can tell.

i used the quadratic formula and got the following roots:

x = 4.302775637732 or x = 0.69722436226801

neither one of them appears to be rational.

you may have meant to solve by squaring.

in that case, you would get:

x^2 - 5x + 3 = (x-2.5)^2 - 2.5^2 + 3 = 0

this becomes (x-2.5)^2 - 6.25 + 3 = 0

this becomes (x-2.5)^2 - 3.25 = 0

this becomes (x-2.5)^2 = 3.25

take the square root of both sides to get:

x-2.5 = plus or minus sqrt(3.25)

this becomes x - 2.5 = plus or mionus 1.802775638

solve for x to get the same thing i got above.

this method is not calloed difference of squares.

it's called completing the square method.

the general principal is.

start with x^2 - 5x + 3 = 0

move the 3 over to the other side of the equation to get:

x^2 - 5x = -3

take half the coefficient of the x term and then square it as shown below:

(x-2.5)^2 - (2.5)^2 = -3

move the - 2.5)^2 to the other side of the equation t0o get:

(x-2.5)^2 = (2.5)^2 - 3

take the square root of both sides of the equation to get:

x-2.5 = plus or minus square root of ((2.5)^2 - 3)

solve for x as i did above.

here's a good reference on completing the square method.

http://www.purplemath.com/modules/sqrquad.htm

here's a good reference on difference of squares method.

http://www.purplemath.com/modules/specfact.htm