Question
$\displaystyle x^2-4=0 $
Attempt
Using the Formula I got 4
Is my answer correct?
From what you learned about factoring quadratics and special factoring formulas, you know that the above can be restated as:
. . . . .$\displaystyle (x\, -\, 2)(x\, +\, 2)\,=\, 0$
What result do you get when you solve the factors?
The Quadratic Formula gives a result (using a = 1, b = 0, and c = -4) that agrees with the "by factoring" result. What formula did you use?
Note: The answer to any "solving" problem can be checked by plugging it back into the original exercise. In your case:
. . . . .$\displaystyle x^2\, -\, 4\, =\, (4)^2\, -\, 4\, =\, 16\, -\, 4\, =\, 12\, \neq \, 0$
no
and the purpose of adding zero to the left side was...?
-4 is your constant term, that is the "c" in your formula. thus you have a = 1, b = 0 and c = -4
applying the quadratic formula, we have
$\displaystyle x = \frac {-0 \pm \sqrt{0^2 - 4(1)(-4)}}{2(1)} = \frac {\pm \sqrt{16}}{2} = \pm \frac 42 = \pm 2$
for problems like these, the quadratic formula is overkill. either do what stapel did, or just add 4 to both sides and then square root both sides (and remember that the right side will have a +/- with it, since there are two square roots to a number)