Question

solve the following quadratic equation
${z}^{4} - 10 {z}^{2} + 9 = 0$
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Answer 1

Let's assume z^2 = x just for the sake of making the equation look less scarier,


${x}^{2} - 10x + 9 = 0 \\ {x}^{2} - 9x - x + 9 = 0 \\ x(x - 9) - (x - 9) = 0 \\ (x - 9)(x - 1) = 0$


Therefore the zeroes are +9 and +1,

Since,
$x = {z}^{2} = 9 \: and \: 1$

$z = \sqrt{9} = + 3 \: and \: - 3$
similarly,

$z = \sqrt{1} = + 1 \: and \: - 1$


So the roots here are -3,-1, 1 and 3.

Answer 2

z^4 - 10z^2 + 9 = 0

z^4 - z^2 - 9z^2 + 9 = 0

z^2(z^2 - 1) - 9(z^2 - 1) = 0

(z^2 - 9)(z^2 -1) = 0

z^2 - 9 = 0               z^2 -1 = 0

z^2 = 9                    z^2 = 1

z = +/-3                    z = +/-1

∴The answers of this bi-quadratic equation are 3, -3, 1, -1

Hope I helped