Question

Practicing using quadratic formula can someone check my work for errors


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

Answer 1

-x^4-x^2+9=0
-1(x^4+x^2-9)=0\[x=\frac{ -b \pm \sqrt{b ^{2}-4ac} }{ 2a }\]
\[x=\frac{ -1\pm \sqrt{1^{2}-4(1)(-9)} }{ 2(1) }\]

Answer 2

\[x=\frac{ -1\pm \sqrt{1-36} }{ 2 }\]

Answer 3

\[x=\frac{ -1\pm \sqrt{-35} }{ 2 }\]

Answer 4

-4(1)(-9) two negatives make a positive

Answer 5

you were solving for x^2
so you need to find for x!

Answer 6

and he is right typo there

Answer 7

You should have noticed something was wrong when you got -35 under the square root. You cannot find the square root of a negative number.

Answer 8

yeah i see that now forgot to change my sign

Answer 9

no you can have negative suare root if you are dealing with complex system

Answer 10

you sill need to solve for x
you have done x^2

Answer 11

what?

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

?

Answer 16

I'm getting confused.

Answer 17

whered you get a -3 ffrom?

Answer 18

I think I did something wrong in the beginning...meh

Answer 19

Hopefully this fixes me and my crazy math

Answer 20

I haven't gotten enough sleep lately. Math at 1 AM does not bode well for me...

Answer 21

\(\normalsize\color{black}{ -x^4+7= x^2-2}\)

\(\normalsize\color{black}{ -x^4+7 \color{red}{-x^2+2}= x^2-2\color{red}{-x^2+2}}\)

\(\normalsize\color{black}{ -x^4+7-x^2+2=0}\)

\(\normalsize\color{black}{ -x^4-x^2+9=0}\)

\(\normalsize\color{black}{ -1(x^4+x^2-9)=0}\)

\(\normalsize\color{black}{ x^4+x^2-9=0}\)

Nowe, let x² = s \(\normalsize\color{black}{ (x^2)^2+x^2-9=0}\)
\(\normalsize\color{black}{ s^2+s-9=0}\)

see what I am doing ?

Answer 22

then you can use the quadratic formula :) Let me know if you need more help.

Answer 23

sorry @SolomonZelman yes i was trying to avoid changing variables because im not eaxactly sure how to put em back so it =x and not s \[\frac{ -1 \pm \sqrt{1^{2}-4(1)(-9)} }{ 2(1)^2}\]

Answer 24

\[\frac{ -1 \pm \sqrt{1+36} }{ 2 }\]

Answer 25

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

Answer 26

and would that be a final answer?

Answer 27

this is you soled for s, getting \(\LARGE\color{black}{ s= \frac{1±\sqrt{37}}{2} }\)

Answer 28

This is just for the s

Answer 29

\(\LARGE\color{black}{ s= \frac{1+\sqrt{37}}{2} }\) and \(\LARGE\color{black}{ s= \frac{1-\sqrt{37}}{2} }\)

but, s= x²

Answer 30

So,

\(\LARGE\color{black}{ x= (~\frac{1± \sqrt{37}}{2}~)^2 }\)

Answer 31

would the 37 stay and sqrt canel and everything else is sqrd?

Answer 32

\[x=\frac{ 2+37 }{ 4 }\] \[x=\frac{ 39 }{ 4 }\]

Answer 33

or stay how you have it

Answer 34

you can rewrite the answer if you want to. You were a bit off when you re-wrote it.


( 1+√37 )² / 2²
( 1+2√37+ 37 ) / 2²
( 2√37+ 38 ) / 4
( √37+ 19 ) / 2 <▬▬

and

( 1-√37 )² / 2²
( 1-2√37 +37) / 2²
( -2√37 +38) / 4
(- √37 +19) / 2

Answer 35

I would put the square roots at the end, although it doesn't matter at all.
like this,
(19 + √37) / 2 and (19 - √37) / 2

in other words,

(19 ± √37) / 2

Answer 36

im a bit confused right now why wouldnt the exponent cancel the sqrt?

Answer 37

you are raising (1+√37) /2 t the second power, not just √37/2 to the second power.

Answer 38

yes everything is being raised to the second power like (1+2)^2
what i dont understand is 1^2=1
37^2=1369
sqrt of 1369=37
2^2=4
wouldnt it be

Answer 39

38/4?

Answer 40

or do you not solve it

Answer 41

you have 2 solutions for s, and they are

\(\Large\color{blue}{\frac{1+\sqrt{37}}{2} }\) and \(\Large\color{blue}{\frac{1-\sqrt{37}}{2} }\)

right ?

Answer 42

yes

Answer 43

sorry this is confusing for me im getting better though :)

Answer 44

So, and we know that s=x², because we substituted s instead of x².

So to find x, we substitute each of the solutions for s, into s=x². Right ?

Answer 45

yes

Answer 46

\(\Large\color{blue}{x^2=\frac{1±\sqrt{37}}{2} }\) Sorry, I was wrong. I don't need to raise both sides to the second power, I take their square root.

Answer 47

\(\Large\color{blue}{x^2=\frac{1±\sqrt{37}}{2} }\)

\(\Large\color{blue}{x^2=\frac{1-\sqrt{37}}{2} }\) and \(\Large\color{blue}{x^2=\frac{1+\sqrt{37}}{2} }\)

\(\Large\color{blue}{x=±\sqrt{ \frac{1-\sqrt{37}}{2} }}\) and \(\Large\color{blue}{x=± \sqrt{ \frac{1+\sqrt{37}}{2} }}\)

Answer 48

ahh i forgot to sqrt it all

Answer 49

no i did that just didnt put in the sqrt sign for 37

Answer 50

Then you would say,

\(\Large\color{blue}{x=\sqrt{ \frac{1-\sqrt{37}}{2} }}\) and \(\Large\color{blue}{x=\sqrt{ \frac{1+\sqrt{37}}{2} }}\)

\(\LARGE\color{blue}{x= \frac{\sqrt{1-\color{white}{s}\sqrt{37}}}{\sqrt{2}} }\) and \(\LARGE\color{blue}{x=\frac{\sqrt{1-\color{white}{s}\sqrt{37}}}{\sqrt{2}} }\)

Answer 51

the last thing, bottom right corner of my last reply, should be +√37

Answer 52

Then multiply top and bottom times √2.
In each solution.

Answer 53

\(\LARGE\color{blue}{x= \frac{\sqrt{2}(\sqrt{1-\color{white}{s}\sqrt{37}}~~)}{2} }\) and \(\LARGE\color{blue}{x=\frac{\sqrt{2} (\sqrt{1+\color{white}{s}\sqrt{37}}~~)}{2} }\)

Answer 54

\(\LARGE\color{blue}{x= \frac{\sqrt{2-\color{white}{s}2\sqrt{37}}}{2} }\) and \(\LARGE\color{blue}{x= \frac{\sqrt{2+\color{white}{s}2\sqrt{37}}}{2} }\)

Answer 55

And each fraction is ± .

Answer 56

so \[x=\sqrt{74}\]\[x=\sqrt{4\sqrt{74}}\]

Answer 57

or keep as fractions?

Answer 58

you are not getting the correct answers. Where do you get √74 from ?
2√37 = √4 × √37 = √148

And this is not necessary. I posted the exact answer and how to obtain it.

Answer 59

oh alright i thought i had to solve for each one im sorry and i got 74 canceling out 2's and multiplying 37 by 2

Answer 60

yeah.... but nothing cancels though.

Answer 61

\(\Large\color{blue}{x=±\sqrt{ \frac{1-\sqrt{37}}{2} }}\)
there should 4 possible answers

Answer 62

you have two that are + or minus
don't forget that