Question

the real and imaginary solution for x^4-32=14x^2

Answer 1

can someone plz help me solve this

Answer 2

It's important to be familiar with equations which may not be quadratic but are quadratic in FORM.
Take this
\[\Large x^4 -32 = 14x^2\]
Certainly, it is quartic (highest exponent is 4) but notice that if you make the substitution \(\large u = x^2\)
\[\Large u^2 - 32 =14u\]
Can you try to solve for u?

Answer 3

give me 5 min

Answer 4

u has two solution u=16
u=-2

Answer 5

That's good. Now we substitute back...
\[\Large u = 16 = x^2\\\Large u = -2 = x^2\]
Now, can you find the four possible solutions for x?

Answer 6

i dont get it

Answer 7

can u plz help

Answer 8

We have u = 16 and u = -2, right?
We undo the substitution and switch back, we now have
\[\Large x^2 = 16 \\\Large \] and \[\Large x^2 = -2\]
Solve both of these...

Answer 9

but how

Answer 10

How? LOL You solved the more difficult \(\large u^2 - 32 =14 u\) but you can't solve \(\Large x^2 = 16 \)
? :3


How about if we bring 16 to the left side... \(\Large x^2 - 16 = 0\)

Answer 11

(x+4)(x-4)=0

Answer 12

And...?
x = ?

Answer 13

-4 and 4

Answer 14

Okay, that's two of them...
now what about \(\large x^2 = -2\) ?

Just get the square root of both sides...

Answer 15

x+4=0
x=-4

Answer 16

^I'm well aware... now get started on
\(\large x^2 = -2 \)

Answer 17

(x-1)(x+2)=0

Answer 18

x=1
x=-2

Answer 19

No... try again... I told you, get the square root of both sides...
and for your information...
\[\Large (x-1)(x+2) = x^2 +x -2\]