help solve

Let z=x^2 --> z^2-5z-36=0 --> (z+4)(z-9)=0 --> z=-4,9 --> x^2=-4 (no real solutions), x^2=9 --> x=3,-3

If you wanted imaginary solutions as well --> x^2=-4 --> x=2i,-2i --> x=3,-3,2i,-2i

What are the solutions of the equation x^4 – 5x^2 – 36 = 0?

Hello Guest!

\( x^4 – 5x^2 – 36 = 0\)

\(u=x^2\)

\( u^2 – 5u – 36 = 0 \)

      p       q

\(u=-\frac{p}{2}\pm\sqrt{(\frac{p}{2})^2-q}\\ u=2.5\pm\sqrt{6.25+36}\\ u=2.5\pm 6.5\)

\(u_1=9\\ u_2=-4\)

\((x_{1,2})^2=9\\ (x_{3,4})^2=-4\)

\(x_1=3\\ x_2=-3\)

\(x_3=+\sqrt{-4}\ complex\ number\\ x_4=-\sqrt{-4}\ complex\ number\)

  !