Question

How do you solve `4x^4-5x^2-9=0`?

Answer 1

`x = \pm 3/2`

Explanation:

Set `t=x^2`. Then, `t^2 = (x^2)^2 = x^4`. The equation becomes

`4t^2-5t-9=0`

Solve it with the usual quadratic formula to obtain `t_1 = -1` and `t_2 = 9/4`

Now we must revert back to `x`:

`t_1 = -1 \implies x^2 = -1`, which is impossible.

`t_2 = 9/4 \implies x^2 = 9/4 \implies x = pm 3/2`

Answer 2

`x=+-3/2" or "x=+-i`

Explanation:

`"make the substitution "u=x^2`

`rArr4u^2-5u-9=0`

`"using the a-c method to factor the quadratic"`

`"the factors of the product "4xx-9=-36`

`"which sum to - 5 are + 4 and - 9"`

`"split the middle term using these factors"`

`4u^2+4u-9u-9=0larrcolor(blue)"factor by grouping"`

`color(red)(4u)(u+1)color(red)(-9)(u+1)=0`

`"take out the "color(blue)" common factor "(u+1)`

`rArr(u+1)(color(red)(4u-9))=0`

`"equate each factor to zero and solve for u"`

`u+1=0rArru=-1`

`4u-9=0rArru=9/4`

`"change u back into terms of x"`

`rArrx^2=9/4rArrx=+-3/2larrcolor(red)"2 real roots"`

`x^2=-1rArrx=+-ilarrcolor(red)"2 complex roots"`