Algebra/Trig problem!?
What number must be added to each side of the equation x^2+3x=0 in order to make the left member the perfect square of a binomial?
PLEASE explain how you got the answer!
What number must be added to each side of the equation x^2+3x=0 in order to make the left member the perfect square of a binomial?
PLEASE explain how you got the answer!
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ax^2 + bx = c
(b/((2)(a))^2 completes the square
In your case
a = 1
b = 3 so
(3/(2)(1))^2 = (3/2)^2 = 9/4
x^2 + 3x + 9/4 = 9/4
The constant comes from above the second to last step in completing the square
(x + 3/2)^2 = 9/4
x + 3/2 = +- 3/2
x = - 3/2 +- 3/2
x = 0 or - 3
this can be checked by
x^2 + 3x = 0
x(x + 3) = 0
x = 0 or x = - 3
(b/((2)(a))^2 completes the square
In your case
a = 1
b = 3 so
(3/(2)(1))^2 = (3/2)^2 = 9/4
x^2 + 3x + 9/4 = 9/4
The constant comes from above the second to last step in completing the square
(x + 3/2)^2 = 9/4
x + 3/2 = +- 3/2
x = - 3/2 +- 3/2
x = 0 or - 3
this can be checked by
x^2 + 3x = 0
x(x + 3) = 0
x = 0 or x = - 3