Question

Ready for a True Challenge? Solve: (9x^2-4)^4-10(9x^2-4)^2+9=0

Answer 1

\[(9x^2 - 4)^4 - 10(9x^2 - 4) 2 + 9 = 0\]
look like that?

Answer 2

yes exactly

Answer 3

actually the last pary should be a Square^2 +9

Answer 4

\[(9x^2 - 4)^4 - 10(9x^2 - 4)^2 + 9 = 0\]
got it

Answer 5

Yes that is it :) thank you!! sorry i am new to this and learning how to write the equations

Answer 6

\[(9x^2 - 4)^4\]
is
\[(9x^2 - 4)(9x^2 - 4)(9x^2 - 4)(9x^2 - 4)\]
right?

Answer 7

yes...thats as far as i got..

Answer 8

At first i thought oh this is easy x is 0 because it seems everything cancels itself out...but sill confused

Answer 9

Well done @Tameiks315

Answer 10

i got it right? lol

Answer 11

I wrote the problem incorrectly it should be \[(9x^2-4)^4-10(9x^2-4)^2+9\]

Answer 12

and yes =0

Answer 13

@lgbasallote please help :)

Answer 14

the first step is to let a = 9x^2 - 4

so your equation becomes

\[a^4 - 10a^2 + 9\]
okay?

Answer 15

ok

Answer 16

now... i let b = a^2

so this becomes \[\huge b^2 - 10b + 9\]
do you follow?

Answer 17

yes you are factoring it out..i tried but definately missed a step

Answer 18

now.... i break down the middle term

\[\huge b^2 - 9b - b + 9\]do you know how to factor this out?

Answer 19

do you mind breaking it down please...

Answer 20

i have several problems such as this so if i can have one as a true worked out example it would help me greatly

Answer 21

let me show you an example

\[\huge x^2 + 2x + 1\]
\[\huge \implies x^2 + x + x + 1\]
\[\huge \implies (x^2 + x) + (x+1)\]
\[\huge \implies x(x + 1) + (x+1)\]
\[\huge \implies (x+1)(x+1)\]
got it now?

Answer 22

i think so...\[(b-3)(b-3)=0\]

Answer 23

this is where i get stuck because b would equal 3?

Answer 24

it would if the answer was (b-3)(b-3) = 0...but it's not

Answer 25

\[\huge (b^2 - 9b - b + 9)\]
\[\huge \implies b(b - 9) - (b-9)\]
so what do you think is the factored form?

Answer 26

i am still stuck :/ b=-9, b=9

Answer 27

what? how did you get that?

Answer 28

what do you get if you factor out b- 9 from what i last wrote?

Answer 29

i totally bypassed the b in fron of the (b-9)-(b-9) for some reason when my screen updated the equations come over funny for some reason it sayd huge and has brackets

Answer 30

break out (9x^2 -4)^2 first.

(81x^4 -72x^2+16)

then i multiplied that by -10

-810x^4+720x^2-160+9
-810x^4+720x^2-151

now, remember that
(9x^2 -4)^2 is (81x^4 -72x^2+16) right?

so square that to get your equation.
(81x^4 -72x^2+16)(81x^4 -72x^2+16)

6561x^8-1164x^6+7776x^4-2304x^2+256

so now you can add that term with your first term just as the equation calls for (9x^2-4)^2+(9x^2-4)^4

6561x^8-1164x^6+7776x^4-2304x^2+256
+ -810x^4+720x^2-151
----------------------------------------
6561x^8-1164x^6+6696x^4-1584x^2+105 = 0

factor
3(3x-1)(3x+1)(3x^2-1)(9x^2-7)(9x^2-5)=0
so
3x-1=0
3x+1=0
3x^2-1=0
9x^2-7=0
9x^2-5=0

thus your answers are
x = -1/3
x=1/3
x=1/sqrt(3)
x=-1/sqrt(3)
and x=sqrt(5)/3

Answer 31

How did you get the 72x?