Question

What are the solutions of the equation x6 – 9x3 + 8 = 0? Use u substitution to solve. x = –1 and x = –2 x = 1 and x = 2

Answer 1

Let u=x^3
the original equation becomes: u^2-9u+8=0
factor this quadratic equation: (u-8)(u-1)=0
u=8, or u=1
that means x^3=8 or x^3=1
x=2  x=1

Answer 2

Answer:

$x=1$ and $x=2$

Step-by-step explanation:

Let

$u=x^{3}$

Remember that

If $u=x^{3}$

then

$u^{2}=x^{6}$

we have

$x^{6} -9x^{3}+8=0$

Substitute

$u^{2} -9u+8=0$

The formula to solve a quadratic equation of the form $ax^{2} +bx+c=0$ is equal to

$x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}$

in this problem we have

$u^{2} -9u+8=0$

so

$a=1\\b=-9\\c=8$

substitute in the formula

$u=\frac{9(+/-)\sqrt{(-9)^{2}-4(1)(8)}} {2(1)}$

$u=\frac{9(+/-)\sqrt{49}} {2}$

$u=\frac{9(+/-)7} {2}$

$u=\frac{9+7} {2}=8$

$u=\frac{9-7} {2}=1$

remember that

$u=x^{3}$

so

for $u=8$

$8=x^{3}$ -------> $x=2$

for $u=1$

$1=x^{3}$ -------> $x=1$