Question

If α and β are the zeroes of the polynomial x² + 5x - 14, find the value of α² + β² (

Answer 1

Answer:

53

Step-by-step explanation:

(α+β)²=α²+β²+2αβ

So

α²+β²=(α+β)²-2αβ

we know that

(α+β)=  -b/a = -5

(α+β)²= 25                                      (1)

and

αβ= c/a = -14

2αβ= -28                                       (2)

now from (1)-(2)

25-(-28)=25+28 = 53

Answer 2

$\huge{\bf{\underline{\underline{\pink{Question}}}}}$

If α and β are the zeroes of the polynomial x² + 5x - 14, find the value of α² + β²

$\huge{\bf{\underline{\underline{\red{ANSWER}}}}}$

$we \: know \: that,$

$( \alpha + \beta )^{2} = \alpha ^{2} + \beta^{2} + 2 \alpha \beta$

$so, \\ { \alpha }^{2} + { \beta }^{2} = ( \alpha+ \beta)^{2} - 2 \alpha \beta$

$we \: also \: know \: that \: for \: sum \: of \: \\ zero \: for \: any \: quadratic \: \\ polynomial : \: ( \alpha + \beta ) = \frac{ - b}{a}$

= -5

${( \alpha + \beta )}^{2} </p><p>= 25...(1)$

$we \: also \: know \: that \: for \: any \: \\ qudratic \: polynomial \: product \: of \\ zero \: is : ( \alpha \beta ) =\frac{ c}{a}$

=-14

$2 \alpha \beta = - 28$.... (2)[/tex]

$now \: from \: 1^{st} and \: 2^{nd}$

$\implies25 - ( - 28) \\ \implies \: \: \: \: 25 + 28 \\ \implies \: \: 53$

$hope \: its \: help \: u.$