Question

Solve for the roots in the following equation. Hint: Factor both quadratic expressions.

(x^4 + 5x^2 - 36)(2x^2 + 9x - 5) = 0

Answer 1

May I help?

Answer 2

go ahead 3mar

Answer 3

The two factors to begin with each can be factorised.

First one, the quartic, best to convert into a quadratic form
Second one, factorise as a quadratic trinomial.

Answer 4

I have no idea what you're talking about. I haven't even been taught this. I basically just get it thrown at me and get crappy "help" from my teacher.

Answer 5

Sorry for late.

Answer 6

Let's begin with left parentheses.

Answer 7

Can you factorize this:
\[x^4+5x^2-36=0\]?

Answer 8

I don't think they know.
For that one best to let u = x^2 and convert it to a quadratic
then x^4 + 5x^2 - 36 becomes u^2 + 5u - 36

Answer 9

@redbirdd, Are you there?

Answer 10

where is she to proceed with you!?

Answer 11

the method itself is not exactly straight forward. Here's a link that can help. The method is known as the 'product, sum factor' (PSF) method.
https://www.mathsisfun.com/algebra/factoring-quadratics.html

Answer 12

It is the rule and it is applicable for all such problems.
\[(x^4+5x^2-36)(2x^2+9x-5)=0\]
\[(x^2+9)(x^2-4)(x+5)(2x-1)=0\]
\[(x^2+9)(x-2)(x+2)(x+5)(2x-1)=0\]

each parentheses is equal to zero,
then you can find the roots:::
\[x^2+9=0\]
\[x^2=-9\]
\[x=\pm \sqrt{-9}=\pm (\sqrt{-1}*\sqrt{9})=\pm 3i\]
where \[i=\sqrt{-1}\] is an imaginary number


\[x-2=0\]
\[x=2\]



\[x+2=0\]
\[x=-2\]



\[x+5=0\]
\[x=-5\]



\[2x-1=0\]
\[x=\frac{ 1 }{ 2 }\]


So the solution is \[{3 , -3 , -5 , \frac{ 1 }{ 2 } }\] if x belongs to real numbers
and it is \[{3 , -3 , -5 , \frac{ 1 }{ 2 } , 3i , -3i}\] for x belongs to imaginary numbers

Answer 13

Thank you for the medal! mmw