Question

Can someone check my work



2. -x^4+7= x^2-2

Answer 1

\[-x ^{4}+7-(x ^{2}-2)=0\]

Answer 2

\[-x ^{4}+7-x ^{2}+2\]

Answer 3

So far so good, except for the "= 0" that disappeared. It is NOT optional.

Answer 4

\[-x ^{4}-x ^{2}+9=0\]

Answer 5

Well? How about \(x^{4} + x^{2} - 9 = 0\)?

Answer 6

now im putting it in u^2=x^4 u=x^2 form
\[-u ^{2}-u+9\]
dividing both sides by -1
\[u ^{2}+u-9\]
add 9 to both sides
\[u ^{2}+u=9\]

Answer 7

That last thing was no good. you need a quadratic equation equal to zero and you just killed it. Go back one step. That's where I left you...

\(u^{2} + u - 9 = 0\)

Now what?

Answer 8

adding (1/2)^2 to both sides
(this is where i got confused)
\[u ^{2}+u+\frac{ 1 }{ 4 }=\frac{ 37 }{ 4 }\]
\[(u+\frac{ 1 }{ 4 })^{2}=\frac{ 37 }{ 4 }\]
\[u+\frac{ 1 }{ 4 }=\frac{ \sqrt{37} }{ 4 }\]
\[u=\frac{ \sqrt{37}-1 }{ 2 }\]

Answer 9

do i leave it like that or subtract 1 into 37 find the sqrt of 36 and divide by 2?

Answer 10

Now, see, you didn't tell me you were "Completing the Square". Fair enough. The quadratic formula is exactly the same thing.

third equation is no good. the right-hand side should be \(\dfrac{\sqrt{37}}{2}\). Start again from there.

Answer 11

Also, that same one needs a \(\pm\) in there, somewhere.

Answer 12

can you explain more please why it wouldnt stay 37/4?

Answer 13

This is correct:

\(\left(u^2 + u +\dfrac{1}{4}\right) = \dfrac{37}{4}\)

The next step was not correct. Almost!

\(\left(u +\dfrac{1}{2}\right)^{2} = \dfrac{37}{4}\)

See that 1/2 in there? You still had 1/4.

The next step was a little off. You found the square root fo the numerator, but not the denominator. You also didn't notice that a negative value would suffice. It should be:

\(u +\dfrac{1}{2} = \pm\sqrt{\dfrac{37}{4}} = \pm\dfrac{\sqrt{37}}{2}\)

Understand? Can you finish?

Answer 14

im completely lost lol sorry from here would i subtract 1/2 from both sides

Answer 15

but i do understand why the 4 changed

Answer 16

The 4 didn't change. (1/2)^2 = 1/4

Answer 17

no the \[\sqrt{\frac{ 37 }{ 4 }}\] went to \[\frac{ \sqrt{37} }{ 2 }\] right?

Answer 18

\(\sqrt{4} = 2\)

Answer 19

yes thats what i was getting at now would i subtract 1/2?

Answer 20

@phi?

Answer 21

Why is there a question? You have (u + 1/2). How else will you be solving for 'u'?

Answer 22

subtracting 1/2

Answer 23

Do it. Where does that lead?

Answer 24

\[u=\frac{ \sqrt{37}-1 }{ 2 }\]

Answer 25

No good. You still missed that positive or negative solutions exist for that square root.

\(u = \dfrac{-1 \pm \sqrt{37}}{2}\)

Answer 26

It looks like you have the right idea, and need to polish up the details

Answer 27

\[u=\frac{ -1\sqrt{37} }{ 2 } u=\frac{ \sqrt{37}-1 }{ 2 }\]

Answer 28

?? That first one doesn't mean anything. Why not write it just like I did? LaTeX \ p m

Answer 29

you left out a minus sign in the first solution

Answer 30

\[u=\frac{ -1-\sqrt{37} }{ 2 }\]

Answer 31

It's okay to write the solutions separately, but you probably don't want to do that. If we were solving for "u", we'd be done. We're supposed to be solving for 'x'.

Answer 32

to finish, remember that the u = x^2
so
\[ x= \pm \sqrt{u} \]
you end up with 4 solutions

Answer 33

this is confusing me

Answer 34

2 with quadratic formula and 2 with completing the square?

Answer 35

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

Answer 36

Just as a note -- Sometimes, and I'm serious, just remembering what it is you are doing is the hardest part of the problem. We've wandered a long way from where we started. Just go back and review. Get it all in your head, again. It will fit! :-)

Answer 37

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

Answer 38

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

Answer 39

\[x=\frac{ 1\pm \sqrt{37} }{ -2 }\]

Answer 40

ty @tkhunny and @phi sorry if im frustrating yall :/

Answer 41

Why make it harder than it has to be? Generally, we want to get that leading coefficient positive.

-x^4+7= x^2-2 ==> -x^4 - x^2 + 9 = 0 ==> x^4 + x^2 - 9 = 0

\(x^{2} = \dfrac{-1\pm\sqrt{1^{2} - 4(1)(-9)}}{2(1)} = \dfrac{-1\pm\sqrt{37}}{2} \rightarrow x = \pm\sqrt{\dfrac{-1\pm\sqrt{37}}{2} }\)

It's not pretty.

Answer 42

I don't get frustrated, but sometimes the asker does.

you can use either the quadratic formula or complete the square to solve a quadratic. You did that for
u^2+u−9=0

but, as you may recall, the question was
x^4+x^2−9=0

where you let x^2 = u
once you find the values for u, you can solve for the values of x
using
\[ x= \pm \sqrt{u} \]

Answer 43

Personally, I am not a fan of the direct substitution. As you noticed, this enticed you to forget the original problem statement! I MUCH prefer just recognizing the Quadratic and laying it out - as demonstrated above.

Answer 44

so with completing the square its \[x^2=\frac{ -1\pm \sqrt{37} }{ 2 }\]
quadratic \[x=\pm \sqrt{\frac{ -1\pm \sqrt{37} }{ 2 }}\]

Answer 45

quadratic seems easier than the first one

Answer 46

\[x=\sqrt{\frac{ -1\pm \sqrt{37} }{ 2 }}\]

Answer 47

yes, with a ± in front
\[ x=\pm \sqrt{\frac{ -1\pm \sqrt{37} }{ 2 }} \]
that is short-hand for 4 separate solutions

Answer 48

ahh alright i have 1 more question can you help me or walk me through with