how would you solve the equation x^4-25x^2+144=0?

Answer 1

Let x^2 = y
so.. y^2 - 25y + 144 = 0.
factorising
(y-9)(y-16) = 0
again put x^2 = y
(x^2-9)(x^2-16) = 0.
this can be written as
(x-3)(x+3)(x-4)(x+4) = 0. as (a^2-b^2) = (a+b)*(a-b)
solution is 3,-3,4,-4

Answer 2

A couple steps of factoring:

x^4 - 25x^2 + 144 = 0

(x^2 )(x^2 )

-16 * -9 =144
-16 + -9 = -25

So, x^4 - 25x^2 + 144 = (x^2 - 9)(x^2 - 25) = 0

(x^2 - 9) = 0 -------> x^2 = 9; x = +3 or -3
or
(x^2 - 25) = 0 ------> x^2 = 25; x = +5 or -5

x = -5, -3, 3, 5

Answer 3

I would solve this by synthetic division..

1 0 -25 0 144
-4
1 -4 -9 36 0

x^3-4x^2-9x+36

1 -4 -9 36
4
1 0 -4 0

x2-4=0

2 and -2 is a root as well as -4 and 4

Answer 4

You can rewrite it like this: Let y = x^2 and
y^2 - 25y + 144 = 0.

Now, are there 2 numbers that multiply to give you 144 and add to give you -25? Yes, -9 and -16.

(y-9)(y-16) = 0

Now, put back x^2 for y and you get:
(x^2-9)(x^2-16) = 0. And this factors further

(x-3)(x+3)(x-4)(x+4) = 0.

So, the values of x that will satisfies this equation are: 3, -3, 4, and -4

Answer 5

You could try synthetic division. Let's try (x - 3).

3 I 1 0 -25 0 144
3 9 -48 -144
1 3 -16 -48 0

That leaves us with (x - 3)(x^3 + 3x^2 - 16x - 48) = 0. Divide again, this time using (x + 3):

-3 I 1 3 -16 -48
-3 0 48
1 0 -16 0

That leaves us with (x - 3)(x + 3)(x^2 - 16) = 0. Further factoring is easy:

(x - 3)(x + 3)(x - 4)(x + 4) = 0

Solution set: x = -4, -3, 3, 4

Answer 6

x=4

Answer 7

Let y = x^2
y^2 - 25y + 144 = 0
(y - 16)(y - 9) = 0
y = 16 or y = 9
x^2 = 16 or x^2 = 9
x = +4, -4, +3, or -3

Answer 8

x^4-25x^2+144=0(1)
t=x^2,T>0
(1) <-> t^2-25t+144=0
<-> t=y1 v t*=y2
if y1 or y2 <0 -> no choice y1 , y2
after find y1, y2 :
x^2=y1
x^2=y2

this is a popular way to solve your equation
you will finh x easily

Answer 9

x^4 - 25x^2 + 144 = 0

x^2 = (-b ± sqrt(b^2 - 4ac))/2a

x^2 = (-(-25) ± sqrt((-25)^2 - 4(1)(144)))/(2(1))
x^2 = (25 ± sqrt(625 - 576))/2
x^2 = (25 ± sqrt(49))/2
x^2 = (25 ± 7)/2
x^2 = (32/2) or (18/2)
x^2 = 16 or 9
x = -4, 4, 3, or -3

ANS : = -4, -3, 3, 4

the factored form would be (x - 4)(x + 4)(x - 3)(x + 3)

Answer 10

take y = x^2
then y^2 - 25y + 144 = 0
or, y^2 - 16y - 9y + 144 = 0
or, y(y - 16) - 9 (y - 16) = 0
or, (y - 9) (y - 16) = 0
or, y = 9 or 16
now replacing x^2 = 9 or 16
or x = 3 or 4