Solve for x: (x^4 - 1) / (x^3) = 0

Question

anonymous · Answer

x=-1
x=1

dumbcow · Answer

denominator is irrelevant, since zero divided by anything is still 0

x^4 - 1 = 0
x^4 = 1
x = +-1

anonymous · Answer

Thanks..can u help me with this: i was trying to solve it using the quadratic formula. im not sure if its rite thou

dumbcow · Answer

sure

anonymous · Answer

solve for x: 2x^2 +5x = 8

dumbcow · Answer

let a = 2, b=5, c = -8
$$x = \frac{-5 \pm \sqrt{(-5)^{2}-4(2)(-8)}}{2(2)}$$

dumbcow · Answer

$$x = \frac{-5 \pm \sqrt{89}}{4}$$

anonymous · Answer

Isnt teh quadratic formula x = -b +/ sqrt b^2 not (-b)^2

anonymous · Answer

so shudnt it be jsut 5^2 not (-5)^2

anonymous · Answer

never mind...i get it cuz it makes no dffnce

anonymous · Answer

teh outcoem is teh same

anonymous · Answer

@arizona cardianl can u help me with a problem..its long and confusing?????

dumbcow · Answer

yesh sure, and you are right i messed up but since its squared it makes no difference
good catch though :)

anonymous · Answer

Solve for x: 
(x+1)^2 (x-2) + (x+1)(x-2)^2 = 0

dumbcow · Answer

factor out what they have in common
each term has a (x+1)(x-2)

$$(x+1)(x-2)((x+1)+(x-2)) = 0$$
$$(x+1)(x-2)(2x-1) = 0$$

anonymous · Answer

oh ok so then teh solutions are x = -1, x = 2, x = 1/2

dumbcow · Answer

yep
make sense now

anonymous · Answer

yes..i thought it was going to be harder because there were exponents.

anonymous · Answer

i forgot absolute values: so the q askes$$\left| x-3 \right|<7 $$
when we solve it do we have to set it to (x-3) is less tahn 7 and teh otehr one to (x-3) greater tahn -7?

dumbcow · Answer

yes that is correct
$$-7 < x-3 < 7$$