Question

$if \: a \: \: and \: \beta \: are \: the \: zeroes \: of \: f(x) = 4x {}^{2} - 5x - 1 \: then \: evaluate \\ (i) \: a {}^{2} + \beta {}^{2} \\ (ii){a}^{3} + { \beta }^{3 } \\ (iii) {a}^{4 } + { \beta }^{4} \\ (iv) {a}^{2} - { \beta }^{2}$
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Answer 1

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Answer 2

Answer:

$\alpha^{2}+\beta^{2}=\dfrac{33}{16} \\ \\ \alpha^{3}+\beta^{3}=\dfrac{185}{64} \\ \\ \alpha^{4}+\beta^{4}=\dfrac{593}{156} \\ \\ \alpha^{2}-\beta^{2}=\dfrac{5\sqrt{33}}{16}$

Step-by-step explanation:

$f(x)=4x^{2}-5x-1$ is the given equation. Its roots are $\alpha$ and $\beta$

For a quadratic equation in 'x':-

Sum of roots, $\alpha +\beta =\dfrac{-(-5)}{4}=\dfrac{5}{4}$

Product of roots, $\alpha \times \beta = \dfrac{-1}{4}$

To evaluate the required expressions:-

$1. \ \ \alpha ^{2} +\beta ^{2} = (\alpha +\beta )^{2}-2\alpha \beta \\ \\ \alpha^{2} +\beta ^{2} = \left( \dfrac{5}{4}\right)^{2}-2\left(\dfrac{-1}{4}\right) \\ \\= \dfrac{33}{16} \\ \\ \\2. \ \ \alpha ^{3} +\beta ^{3} = (\alpha +\beta )^{3}-3(\alpha +\beta)(\alpha \beta ) \\ \\ \alpha^{3} +\beta ^{3} = \left(\dfrac{5}{4}\right)^{3}-3 \ \left(\dfrac{5}{4}\right)\left(\dfrac{-1}{4}\right) \\ \\= \dfrac{185}{64}$

$3. \ \ \alpha ^{4} +\beta ^{4} = (\alpha +\beta )^{4}-2(\alpha\beta)^{2} \\ \\ \alpha^{4} +\beta ^{4} = \left(\dfrac{5}{4}\right)^{4}-2\left(\dfrac{-1}{4}\right)^{2} \\ \\= \dfrac{625}{256}- \dfrac{2}{16}\\ \\=\dfrac{593}{256} \\ \\ \\4. \ \ \ \alpha ^{2}-\beta^{2}=(\alpha +\beta )(\alpha -\beta ) \\ \\= \left(\dfrac{5}{4}\right) \times \dfrac{\sqrt{41}}{4} \\ \\ \\=\dfrac{5\sqrt{41}}{16}$

Note:-

To calculate $(\alpha -\beta )$ use $(\alpha -\beta )^{2}$