Question

if alpha and beta are zeros of the equation 6x^2+x-2=0.find alpha divide by beta +beta divide by alpha. if one of the zero of the polynomial 2x^2+4=0 is R. find the other zero also find the value of P​.

please please please answer me soon. I need help of you guys.

Answer 1

$\underline{\underline{ \bf{Answer}}} : - \\ \implies \: - \frac{25}{12} \\ \\ \underline{\underline{\bf{Explanation}}} : -$

According to the question→

$\bf{\alpha \: and \: \beta \: are \: zeros \: of \: equation} \: \\ \\ \bf{ 6 {x}^{2} + x - 2 = 0}$

We know that,

$\bf{sum \: of \: zeros \: ( \alpha + \beta ) = \frac{ - coefficient \: of \: x}{coefficient \: of\: {x}^{2} } } \\ \\ \bf{ ( \alpha + \beta ) = \frac{ - 1}{6} }$

And ,

$\bf{product \: of \: roots \: ( \alpha \beta ) = \frac{constant \: term}{coefficient \: of \: {x}^{2} } } \\ \\ \bf{ \alpha \beta = \frac{ - 2}{6} }$

Now ,

We find →

$\bf{ \implies \: \frac{ \alpha }{ \beta } + \frac{ \beta }{ \alpha } } \\ \\ \bf{ \implies \: \frac{ { \alpha }^{2} + { \beta }^{2} }{ \alpha \beta } } \\ \\ \implies \: \bf{ \frac{ {( \alpha + \beta )}^{2} - 2 \alpha \beta }{ \alpha \beta } }$

Now ,Given above ,

$\implies \: \frac{( { \frac{ - 1}{6}) }^{2} }{ \frac{ - 2}{6} } - 2 \\ \\ \implies \: \frac{1}{36} \times \frac{ - 6}{2} - 2 \\ \\ \implies \: \frac{ - 1}{12} - 2 = - \frac{25}{12}$