Question

If a and ß be the zeroes of the polynomial 2x² + 3x – 6, find the value of
$\alpha \div \beta + \beta \div \alpha$
​

Answer 1

Step-by-step explanation:

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Answer 2

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$2x {}^{2} + 3x = 6 \\ \alpha + \beta = \frac{ - b}{a} \\ \;\large{\boxed{\bf{\red{\alpha + \beta = \frac{ - 3}{2}}}}} \\ \\ \alpha \times \beta = \frac{c}{a } \\ \alpha \times \beta = \frac{ - 6}{2} \\ \;\large{\boxed{\bf{\red{\alpha \times \beta = - 3}}}}$

$\frac{ \alpha }{ \beta } + \frac{ \beta }{ \alpha } \\ \\ \frac{ \alpha {}^{2} + \beta {}^{2} }{ \alpha \beta } \\ \\ \sf\blue{Using\: Formula ..\: } \\ \\ ⇒ \frac{( \alpha + \beta ) {}^{2} - 2 \alpha \beta }{ \alpha \beta } \\ ⇒ \frac{( \frac{ - 3}{2}) {}^{2} - 2( - 3) }{ - 3} \\ ⇒ \frac{ \frac{9}{4} + 6}{ - 3} \\ ⇒ \frac{33}{4 \times - 3} = \sf\red{\frac{ - 11}{4}}$

$\;\large{\boxed{\bf{\purple{Answer \: is \: \frac{ - 11}{4} }}}}$

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