Question

Match the equation with its solution.

x4 -13x2 + 36 =0

x4 - 11x2 + 24 =0

x - 7 \sqrt{x} =-12

x - 5 \sqrt{x} +6=0

x4 + x2 =12

Answer 1

don't see any solutions...

Answer 2

Well, here's a hint: anything with an \(x^4\) in it here will have 4 solutions. Everything else will have 2 solutions.

Answer 3

You can try factoring, or try out the possible answers in an equation and see if they work.

Answer 4

Do you know the rational root/zero theorem?

Answer 5

no

Answer 6

Okay, well, I won't suggest you use it then :-)

Let's work through the first one as an example:

\[x^4-13x^2+36=0\]Could you factor that if it was\[x^2-13x+36=0\]?

Answer 7

you need to find a pair of numbers that multiply to 36, and add to -13

Answer 8

-4 and 9

Answer 9

how about -4 and -9? That would be \[ (x-4)(x-9) = x^2-9x-4x+36 = x^2-13+36\]
Well, the only difference between that and what you have is the exponents are twice as big here. It factors the same way:
\[(x^2-4)(x^2-9) = x^4-9x^2-4x^2+36 = x^4-13x^2+36\]
But wait, notice that both of the factors can be factored again, as the difference of squares:
\[x^2-4 = (x-2)(x+2)\]\[x^2-9 = (x-3)(x+3)\]That means that \[(x-2)(x+2)(x-3)(x+3) = x^4-13x^2+36=0\]And what are the solutions to all of those product terms?

Answer 10

(-4 and 9 don't work, has to be -4 and -9)

Answer 11

ok

Answer 12

x-2 = 0, x =
x+2 = 0, x =
x-3 = 0, x =
x+3 = 0, x =

Answer 13

x=2
x=-2
x=3
x=-3

Answer 14

Right! there's a shorthand sometimes used
\(x = \pm 2, \pm 3\) which means the same thing as your list

Answer 15

and they use a word version of it in the answer list

Answer 16

umm the answer is c

Answer 17

why would you say that?

Answer 18

ummmmmm idk its d

Answer 19

Come on, you're guessing, and there's no need to guess when you have the answer right in front of you. Only one of those answer choices gives you -2 +2 -3 and +3

Answer 20

do you see any square root signs in our answer?

Answer 21

ohhhhh the answer is E and no

Answer 22

that's better :-)

Answer 23

okay, I'll help you with one more of the other kind:

\[x-7\sqrt{x} = -12\]Subtract x from both sides
\[-7\sqrt{x} = -12-x\]Square both sides
\[(-7\sqrt{x})^2 = (-12-x)(-12-x)\]\[49x^2 = 144+12x+12x+x^2\]\[49x^2=144+24x+x^2\]Subtract \(49x^2\) from both sides
\[0=144+24x-48x^2\]Factor or complete square or use quadratic formula

Answer 24

???????

Answer 25

Whoops, sorry, that should be

\[49x = 144+24x + x^2\]
\[0=144-25x+x^2\]

Answer 26

What's your question?

Answer 27

\[x^2-25x+144=0\]

Two numbers that multiply to 144 and sum to -25: -9, -16

\[(x-9)(x-16)=0\]\[x=\]

Answer 28

I have to go, good luck with the others. You should be able to do them by following these examples.