# find real solutions for equation

• Oct 27th 2008, 04:16 AM
jvignacio
find real solutions for equation
hi guys have problems with this question
find real solutions for:

$
x^2 - 2x - 48 = 0
$

i tryed using quadratic but it ends up being a square root minus which is ERROR?

any help? cheers
• Oct 27th 2008, 04:25 AM
mr fantastic
Quote:

Originally Posted by jvignacio
hi guys have problems with this question
find real solutions for:

$
x^2 - 2x - 48 = 0
$

i tryed using quadratic but it ends up being a square root minus which is ERROR?

any help? cheers

The quadratic formula is unnecessary since $x^2 - 2x - 48 = (x - 8)(x + 6) \, ....$

Note: Without seeing your working it's impossible to know what mistake(s) you've made in applying the quadratic formula. At a guess, I'd say you found the discriminant to be either 4 - 4(48) = -188 or -4 - 4(48) = -196 ...... both are wrong. It's (-2)^2 - 4(-48) = 4 - 4(-48) = 196.
• Oct 27th 2008, 04:39 AM
jvignacio
Quote:

Originally Posted by mr fantastic
The quadratic formula is unnecessary since $x^2 - 2x - 48 = (x - 8)(x + 6) \, ....$

Note: Without seeing your working it's impossible to know what mistake(s) you've made in applying the quadratic formula. At a guess, I'd say you found the discriminant to be either 4 - 4(48) = -188 or -4 - 4(48) = -196 ...... both are wrong. It's (-2)^2 - 4(-48) = 4 - 4(-48) = 196.

i get ya. so the answer would be

x = 8 , x = -6 ?
• Oct 27th 2008, 04:41 AM
mr fantastic
Quote:

Originally Posted by jvignacio
i get ya. so the answer would be

x = 8 , x = -6 ?

Yes.