Question

$\displaystyle x^2-4=0 $

Attempt

Using the Formula I got4

Is my answer correct?

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- Apr 5th 2009, 10:16 AMmj.alawamisolution of the quadratic equation
**Question**

$\displaystyle x^2-4=0 $

**Attempt**

Using the Formula I got**4**

Is my answer correct? - Apr 5th 2009, 10:23 AMstapel
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**? (Wink)

**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:

. . . . .$\displaystyle x^2\, -\, 4\, =\, (4)^2\, -\, 4\, =\, 16\, -\, 4\, =\, 12\, \neq \, 0$

:D - Apr 5th 2009, 10:27 AMmj.alawami
- Apr 5th 2009, 10:31 AMmj.alawami
- Apr 5th 2009, 10:34 AMJhevon
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)