Question

What are the solutions of the equation x4 - 9x2 + 8 = 0? Use u substitution to solve.

A. x = 1 and x = 2sqrt2
B. x=+/-1 and x = +/-2sqrt2
C. x=+/-i and x = +/-2isqrt5
D. x = +/-i and x=2sqrt2

Answer 1

`\large\boxed{B.\ x=\pm1\ and\ x=\pm2\sqrt2}`

Explanation:

`x^4-9x^2+8=0\\\\x^(2\cdot2)-9x^2+8=0\qquad\text{use}\ (a^n)^m=a^(nm)\\\\(x^2)^2-9x^2+8=0\\\\\text{substitute}\ x^2=t\geq0\\\\t^2-9t+8=0\\\\t^2-t-8t+8=0\\\\t(t-1)-8(t-1)=0\\\\(t-1)(t-8)+0\iff t-1=0\ \vee\ t-8=0\\\\t-1=0\qquad\text{add 1 to both sides}\\t=1\geq0\\\\t-8=0\qquad\text{add 8 to both sides}\\t=8\geq0`

`t=x^2\to x^2=1\ \vee\ x^2=8\\\\x^2=1\Rightarrow x=\pm\sqrt1\to x=\pm1\\\\x^2=8\Rightarrow x=\pm\sqrt8\to x=\pm√(4\cdot2)\to x=\pm\sqrt4\cdot\sqrt2\to x=\pm2\sqrt2`

Answer 2

Explanation:

Given is the equation of 4th degree in x,

`x^4 - 9x^2 + 8 = 0`

Substitute
`x^2=u`

`u^2-9u+8=0\\(u-1)(u-8)=0`

u=1 and u =8

i.e.
`x^2=1\\x^2 =8`

Solving we get

`x=1,-1,2√(2), -2√(2)`

Option B is right.