Solve x^4-8x^2-9=0

Pls help

Step-by-step explanation: x can be 3, -3, i, or -i.

Explanation:

If you can't find the factoring by looking at this, simplify the equation.

let

x

2

=

y

to make things easier to see

Now we have

y

2

−

8

y

−

9

=

0

See it now?

We can factor into

(

y

−

9

)

(

y

+

1

)

Now substitute back in

x

2

for

y

.

(

x

2

−

9

)

(

x

2

+

1

)

Since

(

x

2

−

9

)

is a difference of two squares,

(

x

−

3

)

(

x

+

3

)

Now we have

(

x

−

3

)

(

x

+

3

)

(

x

2

+

1

)

x can be 3, -3 for the first two parts

x

2

+

1

=

0

can become

x

2

=

−

1

Taking the positive and negative root means

x

=

±

√

−

1

Thus

x

=

±

i

in addition to 3 and -3.

problem:

f(x) = - 2/3x + 5; f(5/2)

substitute 5/2 in for x:

f(5/2) = -2/3(5/2) + 5

simplify the right side:

f(5/2) = -5/3 + 5

f(5/2) = 10/3

f(5/2) = 10/3

i hope this !