Question

Help with solving for x in quadratic equation..

Answer 1

so i converted it to \[y^2-2y-24\] and got y=6 and -4

Answer 2

now how do i bring it back to x?

Answer 3

i set y=x^1/3

Answer 4

im using substitution, is what it's called...i think

Answer 5

y=x^1/3
cubing both sides
y^3 = x
got this ?

Answer 6

ya, so i just square the answers for y= ?

Answer 7

not square, cube them

Answer 8

oh duh, got mah terms mixed up

Answer 9

so x=36 and 16?

Answer 10

o my

Answer 11

sry

Answer 12

216 and -64

Answer 13

remember that \(\huge \sqrt[3]{x}=-4\)
will not have any solutions.
yes, so x is just 216.

Answer 14

how does it not have a solution for \[\sqrt[3]{x}=-4\]?

Answer 15

**any real solution
ok,
say you got -64
\(\sqrt[3]{-64}=\sqrt[3]{64}\times \sqrt[3]{-1}=4\times \sqrt[3]{-1}\)
now that is an imaginary number , so there will be no real solution to that.

Answer 16

in real domain, \(\sqrt[3]x\) will never be negative.

Answer 17

i thought that was only with \[\sqrt{x} \]

Answer 18

You are correct. A third root can be negative. since (-4)^3=-64

Answer 19

its with every \(x^{(1/n)}\)
where n is positive.

Answer 20

x= -64 will not satisfy original equation.

Answer 21

kk

Answer 22

if i plug in -4*-4*-4 into a calculator it comes out as -64

Answer 23

yes, it will,
\(\sqrt[3]x=y\)
and \(x=y^3\)
are different equations,
first one might have one or no solution,
2nd one will have atleast one real solution.

Answer 24

and we are solving here first equation, not the 2nd one...

Answer 25

\[(-64)^\frac{2}{3}-2(-64)^{\frac{1}{3}}-24=0\]

Answer 26

http://www.wolframalpha.com/input/?i=%28-64%29%5E%282%2F3%29-2%28-64%29%5E%281%2F3%29-24%3D0

Answer 27

\[16+8-24=0\]

Answer 28

(-64)^(1/3) is imaginary.

Answer 29

sorry, complex.

Answer 30

http://www.wolframalpha.com/input/?i=%28-64%29%5E%281%2F3%29

Answer 31

both (-64)^(2/3)and (-64)^(1/3) are complex.

Answer 32

so here, you are only left with x=216

Answer 33

thx

Answer 34

wlcm