Question

If alpha are the zeros of PX x square - 5 x + 6 then find one upon alpha plus one upon beta

Answer 1

Answer:

See attachment for solution

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Answer 2

COMPLETE QUESTION :

If alpha and beta are the zeros of P(x),x square - 5 x + 6 then find one upon alpha plus one upon beta.

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{\frac{1}{\alpha}+\frac{1}{\beta}=\frac{5}{6}}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green {\underline \bold{given: }} \\ \tt{: \implies {x}^{2} - 5x + 6 = 0} \\ \\ \tt{: \implies \alpha \: and \: \beta \: are \: the \: zeroes} \\ \\ \red{\underline \bold{to \: find: }} \\ \tt{: \implies \frac{1}{\alpha } + \frac{1}{ \beta } =?}$

• According to given question :

$\tt{: \implies {x}^{2} - 5x + 6 = 0} \\ \\ \bold{solving \: by \: middle \: term \: spliting} \\ \tt{: \implies {x}^{2} - 2x - 3x + 6 = 0} \\ \\ \tt{: \implies x(x - 2) - 3(x - 2) = 0} \\ \\ \tt{: \implies (x - 3)(x - 2) = 0} \\ \\ \green{\tt{: \implies x = 3 \: and \: 2}} \\ \\ \green{\circ \: \alpha = 3} \\ \\ \green{ \circ \: \beta = 2} \\ \\ \bold{for \: finding \: value : } \\ \tt{: \implies \frac{1}{\alpha} + \frac{1}{ \beta } } \\ \\ \tt{: \implies \frac{1}{3} + \frac{1}{2}} \\ \\ \tt{: \implies \frac{2+3}{6} } \\ \\ \green{\tt{: \implies \frac{5}{6} }}$