Question

Solve: (x2 - 3)2 + (x2 - 3) - 2 = 0

Answer 1

\[(x^2-3)^2+x^2-3 - 2 = 0\]
Substitute \[x^2-3=y\]

Thus eq. becomes

\[y^2-y-2 = 0\]
\[=> y^2 - 2y+y-2 = 0\]
\[=>y(y-2)-1(y-2) = 0\]
\[=>(y-1)(y-2)=0\]
\[=> y = 1\] or \[y = 2\]
\[=> x^2-3x = 1\] or \[x^2- 3x = 2\]

Now solve these 2 eq.. to get values pf x

Answer 2

ok thanks for the help

Answer 3

\[x^2 - 3x -1 = 0\] has 2 roots
\[1/2 (3+\sqrt{13})\]

\[1/2 (3-\sqrt{13})\]
Similarly \[x^2 - 3x - 2 = 0\] has 2 roots
\[1/2 (3-\sqrt{17})\]

\[1/2 (3+ \sqrt{17})\]

Answer 4

These are your roots