Question

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

Answer 1

Answer:

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

Step-by-step explanation:

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

$t=x^2\to x^2=1\ \vee\ x^2=8\\\\x^2=1\Rightarrow x=\pm\sqrt1\to x=\pm1\\\\x^2=8\Rightarrow x=\pm\sqrt8\to x=\pm\sqrt{4\cdot2}\to x=\pm\sqrt4\cdot\sqrt2\to x=\pm2\sqrt2$

Answer 2

Answer:

Step-by-step 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

$x=1,-1,2\sqrt{2}, -2\sqrt{2}$

Option B is right.