If α and β are the zeros of the quadratic polynomial f(t)

Question:

If $\alpha$ and $\beta$ are the zeros of the quadratic polynomial $t(t)=t^{2}-4 t+3$, find the value of $\alpha^{4} \beta^{3}+\alpha^{3} \beta^{4} .$

Solution:

Since $\alpha$ and $\beta$ are the zeros of the quadratic polynomial $p(y)=t^{2}-4 t+3$

$\alpha+\beta=\frac{-\text { Coefficient of } x}{\text { Coefficient of } x^{2}}$

$=\frac{-(-4)}{1}$

= 4

$\alpha \beta=\frac{\text { Constant term }}{\text { Coefficient of } x^{2}}$

$=\frac{3}{1}$ 

= 3

We have $\alpha^{4} \beta^{3}+\alpha^{3} \beta^{4}$

$\alpha^{4} \beta^{3}+\alpha^{3} \beta^{4}=\alpha^{3} \beta^{3}(\alpha+\beta)$

$\alpha^{4} \beta^{3}+\alpha^{3} \beta^{4}=(\alpha \beta)^{3}(\alpha+\beta)$

$\alpha^{4} \beta^{3}+\alpha^{3} \beta^{4}=(3)^{3}(4)$

$\alpha^{4} \beta^{3}+\alpha^{3} \beta^{4}=27 \times 4$

$\alpha^{4} \beta^{3}+\alpha^{3} \beta^{4}=108$

Hence, the value of $\alpha^{4} \beta^{3}+\alpha^{3} \beta^{4}$ is 108 .

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