Question

P(x) =2x square-4x+5 If alpha and bita are zeros of p(x) find 1/alpha +1/bita

Answer 1

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

$\green{\tt{\therefore{\frac{1}{\alpha}+\frac{1}{\beta}=\frac{4}{5}}}}$

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

$\green{\underline \bold{Given : }} \\ \tt: \implies p(x) = {2x}^{2} - 4x + 5 \\ \\ \tt: \implies \alpha \: and \: \beta \: are \: zeroes \\ \\ \red{\underline \bold{To \: Find: }} \\ \tt: \implies \frac{1}{ \alpha } + \frac{1}{ \beta } = ?$

• According to given question :

$\tt \circ \: {2x}^{2} -4x + 5 = 0 \\ \\ \tt \circ \: a = 2 \: \: \: \: \: \: b = - 4 \: \: \: \: \: \: c = 5 \\ \\ \bold{For \: sum \: of \: zeroes} \\ \tt: \implies \alpha + \beta = - \frac{b}{a} \\ \\ \tt: \implies \alpha + \beta = \frac{ - ( - 4)}{2} \\ \\ \green{\tt: \implies \alpha + \beta = 2} \\ \\ \bold{For \: product \: of \: zeroes} \\ \tt: \implies \alpha \beta = \frac{c}{a} \\ \\ \green{\tt: \implies \alpha \beta = \frac{5}{4} } \\ \\ \bold{For \: finding \: value} \\ \tt: \implies \frac{1}{\alpha } + \frac{1}{\beta } \\ \\ \tt: \implies \frac{ \alpha + \beta }{ \alpha \beta } \\ \\ \tt: \implies \frac{2}{ \frac{5}{2} } \\ \\ \green{\tt: \implies \frac{4}{5} } \\ \\ \green{\tt \therefore \frac{1}{ \alpha } + \frac{1}{ \beta } = \frac{4}{5} }$

Answer 2

Aɴꜱᴡᴇʀ

➜$\large{ \sf{ \green{ \frac{1}{ \alpha } + \frac{1}{ \beta } = \frac{4}{5} }}}$

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Gɪᴠᴇɴ

$\large \tt{} \alpha \: and \: \beta \: are \: the \: 0s \: of \: p(x) = 2{x}^{2} - 4x + 5$

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ᴛᴏ ꜰɪɴᴅ

$\large\tt{} value \: of \: \frac{1}{ \alpha } \: and \: \frac{1}{ \beta }$

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Sᴛᴇᴘꜱ

First let's find their relationship between their coefficients,

Let's first compare it with the general form ax²+bx+c.

☞ a = 2

☞ b = (-4)

☞ c = 5

$\tt \leadsto{} \alpha + \beta = \large \frac{ - b}{a} \\ \\ \tt{} \leadsto{} \alpha + \beta = \large \cancel \frac{ - ( - 4)}{2} \\ \\ \tt{} \hookrightarrow \red{ \alpha + \beta = 2}$

$\tt \mapsto \alpha \beta = \large{} \frac{c}{a} \\ \\ \tt \mapsto { \orange{\alpha \beta = \large \frac{ 5}{2} }}$

$\rm{} {}so \: now \: \large\frac{1}{ \alpha } + \frac{1}{ \beta } \: is \: equal \: to \\ \\ \tt{} \dashrightarrow{} \frac{ \alpha + \beta }{ \alpha \beta } \\ \\ \tt \dashrightarrow{} \large \frac{ 2}{ \frac{5}{2}} \\ \\ \tt \dashrightarrow{} \large 2 \times \frac{2}{5} \\ \\ \tt \pink { \dashrightarrow{ \large{} \frac{4}{5} }}$

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