Question

if alpha and beeta are zeroes of 2y^2-3y+1 then find alpha by beeta plus beeta plus alpha

Answer 1

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Answer 2

➡HERE IS YOUR ANSWER⬇

■] FORMULA :

Let us consider a quadratic equation in y :

ay² + by + c = 0

If α and β are the roots of this equation, then

α + β = -b/a and αβ = c/a

■] ANSWER :

The given equation is :

2y² - 3y + 1 = 0

If α and β are roots of this equation, then

α + β = -(-3)/2 = 3/2 and αβ = 1/2.

Now,

$\frac{ \alpha }{ \beta } + \frac{ \beta }{ \alpha } \\ \\ = \frac{ { \alpha }^{2} + { \beta }^{2} }{ \alpha \beta } \\ \\ = \frac{ {( \alpha + \beta )}^{2} - 2 \alpha \beta }{ \alpha \beta } \\ \\ = \frac{ { (\frac{3}{2} )}^{2} - 2 \times (\frac{1}{2} ) }{ \frac{1}{2} } \\ \\ = (\frac{ \frac{9}{4} - 1}{ \frac{1}{2} } ) \\ \\ = \frac{ \frac{9 - 4}{4} }{ \frac{1}{2} } \\ \\ = \frac{ \frac{5}{4} }{ \frac{1}{2} } \\ \\ = \frac{5}{4} \times \frac{2}{1} \\ \\ = \frac{5}{2}$

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