Question

@sasogeek please point out any errors you see please :)

-x^4-x^2+9=0

Answer 1

\[x=\frac{ -b \pm \sqrt{b ^{2}-4ac} }{ 2a }\]
a=-x^4 b=x^2 c=9

Answer 2

the discriminant is b^2-4ac.
So, the discriminant is 12^2-4(4)(9)
Which is 144-144
And comes out to be 0.

Answer 3

i think

Answer 4

first ima multiply the equation by -1 to make the variables positive then change the variable to u
u^2=x^4 u=s^2

Answer 5

so
-1(-x^4-x^2+9)
x^4+x^2-9=0

x^4=u^2
x^2=u
so now the equation is
\[u^2+u-9=0\]

Answer 6

a=u^2 b=u c=-9

Answer 7

ok my work here is done

Answer 8

or 1,1,-9 forgot to substitute in the 1's

Answer 9

\[x=\frac{ -1\pm \sqrt{1^2}-4(1)(-9) }{ 2(1) }\]
\[x=\frac{ -1 \pm \sqrt{1+36} }{ 2 }\]

Answer 10

\[x=\frac{ -1 \pm \sqrt{37} }{ 2 }\]
This is all of this and btw the first equation in the last reply is suppose to be sqrt over all but -1 +- on top row

Answer 11

you're making sense for the most part but idk for sure if you're fully correct. i've lost touch with a lot of my math cos my academic career has taken a different path. i believe you're on the right path though. I'll ask a friend to confirm your work :)

Answer 12

now i just solved for x and need x^2 also
\[x=\frac{ -1+\sqrt{37} }{ 2 }\] and \[x=\frac{ -1+\sqrt{37} }{ 2 }\]

Answer 13

okay thanks

Answer 14

ima continue to work so i can get u^2 this is where i confuse myself at times

Answer 15

so those problems are u which is x^2 and we need u^2 also or x^4 to do that im thinking raise everything to the 2nd power \[x=\frac{ -2\pm \sqrt{74} }{ 4 }\] so the last 2 solutions are
\[(x=\frac{ -1 \pm \sqrt{37}}{ 2 })^2\]\[x^2=\frac{ 1 \pm 37 }{ 4 }\]
\[x^2=\frac{ 38 }{ 4 }\]
???

Answer 16

-x^4-x^2+9=0
multiply throughout by -1:
x^4 + x^2 - 9 = 0
Let u = x^2. Then u^2 = x^4
u^2 + u - 9 = 0
a = 1, b = 1, c = -9
plug it into the quadratic formula and solve for u.
Then take the square root to solve for x.

Answer 17

i know all that im just practicing so i dont get as confused while solving

Answer 18

and would like to know when i mess up

Answer 19

your work for solving \(\large u\) looks good and the answer is correct !
\[\large \begin{align}\\u =x^2 &= \dfrac{1}{2}(-1\pm\sqrt{37}) \\~\\ x&= \pm \sqrt{ \dfrac{1}{2}(-1\pm\sqrt{37})}\end{align}\]

you may simplify further if possible or leave it here...

Answer 20

can you explain how you got the 1/2 please? did you multiply from the denominator?

Answer 21

that expression is same as yours, see if this looks better :

\[\large \begin{align}\\u =x^2 &= \dfrac{-1\pm\sqrt{37}}{2} \\~\\ x&= \pm \sqrt{ \dfrac{-1\pm\sqrt{37}}{2}}\end{align} \]

Answer 22

\[\large \dfrac{-1\pm\sqrt{37}}{2} \]

is same as

\[\large \dfrac{1}{2}(-1\pm\sqrt{37}) \]

Answer 23

oh i see now thanks and that would be the final answer? no need to go any further?

Answer 24

I prefer the latter form thats ll... :)

Answer 25

Looks good to me !

Answer 26

unless if there is a way to simplify the radical :
\[\large x= \pm \sqrt{ \dfrac{-1\pm\sqrt{37}}{2}}\]

sometimes we can simplify radicals like these, but i don't think it is possible here... so you're done mostly !

Answer 27

did i help

Answer 28

ok thanks ganeshie :)

Answer 29

There are 4 roots to the original equation: two real and two complex.
The above is a compact way of expressing the 4 roots.