SOLUTION: Is it all equation that can be cross checked becos am confuse when i solve this equation "4x^2+8x-3=0" using completing the square method and i get 1 or-3. But i d LHS is nt=RHS. W

Question 894615: Is it all equation that can be cross checked becos am confuse when i solve this equation "4x^2+8x-3=0" using completing the square method and i get 1 or-3. But i d LHS is nt=RHS. WHY?
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
4x^2 + 8x - 3 = 0

using complete the square method, we get:

4x^2 + 8x = 3

divide both sides of this equaton by 4 to get:

x^2 + 2x = 3/4

complete the square to get:

(x + 1)^2 = 1 + 3/4

simplify to get:

(x + 1)^2 = 7/4

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

x + 1 = plus or minus sqrt (7/4)

subtract 1 from both sides of the equation to get:

x = -1 plus or minus sqrt(7/4)

that should be your answer.

let me confirm for you using my calculator.

the colution is confirmed as good.

your LHS is not equal to your RHS because you messed up somewhere in the factoring when you used the completing the square method.

in order to use the completing the square method, the a term has to be equal to 1.

ax^2 + bx + c = 0 has to be converted to x^2 + *b/a)*x + c/a = 0

if your didn't do that you did it wrong.

you could also have solved this using the quadratic formula which is always reliable and actually easier to use than the completing the square method although it is really good to know how to use the completing the square method because it comes up later on when converting equations from one form to the other.

the quadratic formula, in this equation would be used as follows.
convert the equation to standard form to get:
4x^2 + 8x - 3 = 0
a = 4
b = 8
c = -3

quadratic formula is x = (-b +/- sqrt(b^2 - 4ac)) / (2a)

this will become x = (-8 +/- sqrt(8^2 - 4*4*-3)) / 8

simplify this to get:

x = (-8 +/- sqrt(112)) / 8 which becomes:

x = -1 +/- sqrt(7*16)/8 which becomes:

x = -1 +/- 4*sqrt(7)/8 which becomes:

x = -1 +/- sqrt(7)/2 which becomes:

x = -1 +/- sqrt(7/4)

the solutions are the same even though i had to do some manipulating to get them to look the same.

whether or not they looked the same, they are equivalent because they will give you the same value of x.

here's a link that talks about how to do the completing the square method.

http://www.regentsprep.org/Regents/math/algtrig/ATE3/quadcompletesquare.htm

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