Question

If alpha and beta are two zeros of 2(x)2-5x+8 find (alpha)2÷beta+(beta)2÷alpha

Answer 1

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Answer 2

Answer :

2x² - 5x + 8

Sum of zeroes = - b/a

α + β = -(-5)/2

α + β = 5/2

Product of zeroes = c/a

αβ = 8/2 = 4

Now,

$\frac{ \alpha {}^{2} }{ \beta } + \frac{ \beta {}^{2} }{ \alpha } \\ \\ \frac{ \alpha {}^{3} + \beta {}^{3}}{ \alpha \beta }$

We know that ;

α³ + β³ = (α + β)[(α + β)² - 3αβ]

⇒ α³ + β³ = 5/2 [ (5/2)² - 3 * 4]

⇒ α³ + β³ = 5/2 [ 25 / 4 - 12]

⇒ α³ + β³ = 5/2 [ 25 - 48/12]

⇒ α³ + β³ = 5/2 * (-23/12)

⇒ α³ + β³ = - 115/24

Now,

$\frac{ \alpha {}^{3} + \beta ^{3} }{ \alpha \beta } \\ \\ \frac{ \frac{ - 115}{24} }{4} \\ \\ \frac{ - 115}{24} \times \frac{1}{4} \\ \\ \frac{ - 115}{96}$

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