how do you factor, 4x^4 - 5x^2 - 9 = 0?

Question

so far i have (2x^2)(2x^2)

Answer 1

4x^4 - 5x^2 - 9 = 0

x = (-b ± sqrt(b^2 - 4ac))/(2a)

x = (-(-5) ± sqrt((-5)^2 - 4(4)(-9)))/(2(4))
x = (5 ± sqrt(25 + 144))/8
x = (5 ± sqrt(169))/8
x = (5 ± 13)/8
x = (-8/8) or (18/8)
x = -1 or (9/4)

so the problem factors to

(x^2 + 1)(4x^2 - 9)

But were not done yet, the problem can still factor to

(x^2 + 1)(2x - 3)(2x + 3)

now if you want to take this further, you would get

(x - i)(x + i)(2x - 3)(2x + 3)

Answer 2

Since the equation is of the form 4x^4 - 5x^2 - 9 =0 where the powers of the x are even, we can make a substitution and then use the quadratic formula.

Let u = x^2

Then 4x^4 - 5x^2 - 9 = 4u^2 - 5u - 9

Then substituting into the quadratic formula we get:

u = (-b±√[b² - 4ac]) / 2a
u = (5±√[(-5)² - 4(4)(-9)]) / 2(4)
u = (5±√[25 + 144]) / 8
u = (5±√169) / 8
u = (5±13) / 8

u = (5+13) / 8 = 2.25
u = (5 - 13) / 8 = -1

Then we substitute the values for u back into the equation for x.

u = x^2
±√u = x

±√2.25 = ±1.5 = x
±√-1 = ±i = x

Therefore, the roots of the equation 4x^4 - 5x^2 - 9 =0 are:

x = +1.5, -1.5, +i, -i

Answer 3

That's a start. The first thing to do is to substitute y = x^2 and solve that. First try factors of 4 (for the first term); you did some of that, but you may need to try 1,4 also. Next try factors of 9, the last term. The possibilities are 1,9 or 3,3 (or something irrational, which one can dig out using the quadratic formula if nothing else works). If we try (y+1)(4y-9), we see that we have a solution. Substituting back, we get (x^2+1)(4x^2-9). The first factor is prime, but the second is not; factor that, and get for a final result (x^2+1)(2x+3)(2x-3).

Answer 4

Think of it first as 4x^2-5x-9=0.

Factored, this is (x+1)(4x-9)=0

Then, plug in x^2 for x.

(x^2+1)(4x^2-9)=0

x^2+1 is not factorable.

so you are left with:

(x^2+1)(2x-3)(2x+3)=0

Answer 5

4x^4 - 5x^2 -9 = 0
4x^4 + 4x^2 - 9x^2 - 9 = 0
4x^2(x^2 + 1) - 9(x^2 + 1) = 0
(4x^2 - 9) (x^2 + 1) = 0
(2x - 3)(2x + 3)(x^2 + 1) = 0

Answer 6

Try (4x^2 - 9)(x^2 + 1). This will get 4x^4 + 4x^2 - 9x^2 - 9. Add like terms and you get 4x^4 - 5x^2 - 9.

Answer 7

let a = x^2 so your new equation will be

4a^2 - 5a -9 =0

then find the roots will be

2 1/4 and -1

discard the -1 root when substituting x^2 =a

so x^2 = 2 1/4

so x = +/- sqrt 2 1/4 = +/- 3/2

Answer 8

if x^2=A, then
4x^4 - 5x^2 - 9 = 0 is 4A^2-5A-9=0
(4A-9)(A+1)=0
(4x^2-9)(x^2+1)=0
(2x+3)(2x-3)(x^2+1)=0

Answer 9

let x^2=x, then 4x^2-5x-9=0

(4x-9)(x+1)=0

Than replace with x^2

(4x^2-9)(x^2+1)=0

Answer 10

tough, it could av been easier if d qst was 4x^4 - 5x^2 + 9