CAIE A-Level Further Math AS 1.1 Roots of Polynomial Equations Questions

$y=x^{2}$ so, $x=y^{\frac{1}{2}}$
Correct substitution.

$y^{2}-y+2 y^{\frac{1}{2}}+5=0$ so, $\left(2 y^{\frac{1}{2}}\right)^{2}=\left(-y^{2}+y-5\right)^{2}$
Obtains an equation which eliminates radicals.

$y^{4}-2 y^{3}+11 y^{2}-14 y+25=0$$\alpha^{2}+\beta^{2}+\gamma^{2}+\delta^{2}=2$

B1FT

4

$\alpha^{2} \beta^{2} \gamma^{2} \delta^{2}=25$

B1FT

$\frac{1}{\alpha^{2}}+\frac{1}{\beta^{2}}+\frac{1}{\gamma^{2}}+\frac{1}{\delta^{2}}=\frac{\alpha^{2} \beta^{2} \delta^{2}+\alpha^{2} \beta^{2} \gamma^{2}+\beta^{2} \gamma^{2} \delta^{2}+\alpha^{2} \gamma^{2} \delta^{2}}{\alpha^{2} \beta^{2} \gamma^{2} \delta^{2}}$
Relates to coefficients.

$\frac{14}{25}$
CAO

3

$$\begin{aligned} \alpha^{4}+\beta^{4}+\gamma^{4}+\delta^{4}= \left(\alpha^{2}+\beta^{2}+\gamma^{2}+\delta^{2}\right)^{2} \quad-2\left(\alpha^{2} \beta^{2}+\alpha^{2} \gamma^{2}+\alpha^{2} \delta^{2}+\beta^{2} \gamma^{2}+\beta^{2} \delta^{2}+\gamma^{2} \delta^{2}\right) = 2^{2}-2(11) \end{aligned}$$
Uses formula for sum of squares or uses original equation.

-18
2

$y=x^{2}$$$\begin{aligned} 6 y^{\frac{3}{2}}+p y-3 y^{\frac{1}{2}}-5=0 \Rightarrow y^{\frac{1}{2}}(6 y-3)=-p y+5 y(6 y-3)^{2}=(-p y+5)^{2} \Rightarrow y\left(36 y^{2}-36 y+9\right)=p^{2} y^{2}-10 p y+25 \end{aligned}$$$36 y^{3}-\left(p^{2}+36\right) y^{2}+(10 p+9) y-25=0$
2a 3