Solve over the real numbers: \(x^4-5x^2+4=0\).
\(x=1,2\)
\(x=\pm2\)
\(x=\pm1,\pm2\)
\(x=\pm1\)
Let \(u=x^2\). Then \(u^2-5u+4=0\).
\((u-1)(u-4)=0\), so \(u=1\) or \(u=4\).
Hence \(x^2=1\) or \(x^2=4\).
So \(x=\pm1,\pm2\).
Solve over the real numbers: \(x^4-13x^2+36=0\).
\(x=2,3\)
\(x=\pm2,\pm3\)
\(x=\pm4,\pm9\)
Let \(u=x^2\). Then \(u^2-13u+36=0\).
\((u-4)(u-9)=0\), so \(u=4\) or \(u=9\).
Hence \(x^2=4\) or \(x^2=9\).
So \(x=\pm2,\pm3\).
Solve over the real numbers: \(x^4+x^2-20=0\).
\(x=\pm\sqrt5,\pm2\)
No real solutions
\(x=\pm4\)
Let \(u=x^2\). Then \(u^2+u-20=0\).
\((u+5)(u-4)=0\), so \(u=-5\) or \(u=4\).
Since \(u=x^2\ge 0\), only \(u=4\) gives real solutions.
Hence \(x=\pm2\).
Solve over the real numbers: \(2x^4-5x^2-3=0\).
\(x=\pm\dfrac{1}{\sqrt2}\)
\(x=\pm\sqrt3\)
\(x=\pm\sqrt3,\pm\dfrac{1}{\sqrt2}\)
Let \(u=x^2\). Then \(2u^2-5u-3=0\).
\((2u+1)(u-3)=0\), so \(u=-\dfrac12\) or \(u=3\).
Only \(u=3\) gives real solutions.
Hence \(x=\pm\sqrt3\).
Solve over the real numbers: \(x^4-10x^2+9=0\).
\(x=\pm1,\pm3\)
\(x=1,3\)
\(x=\pm3\)
Let \(u=x^2\). Then \(u^2-10u+9=0\).
\((u-1)(u-9)=0\), so \(u=1\) or \(u=9\).
Hence \(x=\pm1,\pm3\).