(v) 4z4−25z2−21=04z^4 - 25z^2 - 21 = 04z4−25z2−21=0 Solution Let u=z2:4u2−25u−21=0Quadratic formula: u=−b±b2−4ac2au=25±625−4(4)(−21)2(4)=25±625+3368=25±9618=25±318u=25+318=568=7oru=25−318=−68=−34Since u=z2:z2=7 ⟹ z=±7z2=−34 ⟹ z=±i32Therefore: z=7,−7,i32,−i32\begin{aligned} & \boxed{\text{Let } u = z^2:} \\ \\ & 4u^2 - 25u - 21 = 0 \\ \\ & \boxed{\text{Quadratic formula: } u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}} \\ \\ & u = \frac{25 \pm \sqrt{625 - 4(4)(-21)}}{2(4)} = \frac{25 \pm \sqrt{625 + 336}}{8} = \frac{25 \pm \sqrt{961}}{8} = \frac{25 \pm 31}{8} \\ \\ & u = \frac{25 + 31}{8} = \frac{56}{8} = 7 \quad \text{or} \quad u = \frac{25 - 31}{8} = \frac{-6}{8} = -\frac{3}{4} \\ \\ & \boxed{\text{Since } u = z^2:} \\ \\ & z^2 = 7 \implies z = \pm \sqrt{7} \\ \\ & z^2 = -\frac{3}{4} \implies z = \pm \frac{i\sqrt{3}}{2} \\ \\ & \boxed{\text{Therefore: } z = \sqrt{7}, -\sqrt{7}, \frac{i\sqrt{3}}{2}, -\frac{i\sqrt{3}}{2}} \end{aligned}Let u=z2:4u2−25u−21=0Quadratic formula: u=2a−b±b2−4acu=2(4)25±625−4(4)(−21)=825±625+336=825±961=825±31u=825+31=856=7oru=825−31=8−6=−43Since u=z2:z2=7⟹z=±7z2=−43⟹z=±2i3Therefore: z=7,−7,2i3,−2i3