# solution of the quadratic equation

• Apr 5th 2009, 11:16 AM
mj.alawami
Question

$x^2-4=0$

Attempt

Using the Formula I got 4

• Apr 5th 2009, 11:23 AM
stapel
Quote:

Originally Posted by mj.alawami
[SIZE=4]Question

$x^2-4=0$

From what you learned about factoring quadratics and special factoring formulas, you know that the above can be restated as:

. . . . . $(x\, -\, 2)(x\, +\, 2)\,=\, 0$

What result do you get when you solve the factors? (Wink)

Quote:

Originally Posted by mj.alawami
Using the Formula I got 4

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:

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

:D
• Apr 5th 2009, 11:27 AM
mj.alawami
Quote:

Originally Posted by stapel
From what you learned about factoring quadratics and special factoring formulas, you know that the above can be restated as:

. . . . . $(x\, -\, 2)(x\, +\, 2)\,=\, 0$

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:

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

:D

I have done
$x^2 -4 +0 =0$
Then I Used the Quadratic Formula and Got 4
• Apr 5th 2009, 11:31 AM
mj.alawami
Quote:

Originally Posted by mj.alawami

I have done
$x^2 -4 +0 =0$
Then I Used the Quadratic Formula and Got 4

So to Finalize the answer is -2

(Clapping)
• Apr 5th 2009, 11:34 AM
Jhevon
Quote:

Originally Posted by mj.alawami
So to Finalize the answer is -2

no

Quote:

Originally Posted by mj.alawami

I have done
$x^2 -4 +0 =0$
Then I Used the Quadratic Formula and Got 4

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

$x = \frac {-0 \pm \sqrt{0^2 - 4(1)(-4)}}{2(1)} = \frac {\pm \sqrt{16}}{2} = \pm \frac 42 = \pm 2$

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)