Question

if alpha and beta are the zeroes of the polynomials x=x²-2x+5, then find the quadratic polynomial whose zeroes are alpha +beta and 1/alpha + 1/beta

Answer 1

$Compare \: given \: polynomial \: x^{2}-2x+5\\with \: ax^{2}+bx+c , we \:get$

$a = 1 , \:b = -2 \:and \: c = 5$

$\alpha \:and \:\beta \:are \: zeroes \:of \:the \\polynomial$

$i ) \alpha + \beta = \frac{-b}{a}$

$= \frac{-(-2)}{1} \\= 2 \:---(1)$

$ii) \alpha \beta = \frac{c}{a}$

$= \frac{5}{1} \\= 5 \:---(2)$

$iii ) If \: \alpha + \beta \: and \: \frac{1}{\alpha} + \frac{1}{\beta} \: are \\ zeroes \: of \: polynomial ,\:then$

$Sum \:of \:the \: zeroes \\= \alpha + \beta+\frac{1}{\alpha} + \frac{1}{\beta} \\= \alpha + \beta+ \frac{\beta+\alpha}{\alpha \beta } \\= 2+ \frac{2}{5} \\=\frac{10+2}{5} \\= \frac{12}{5} \: --(3)$

$Product \:of \:the \: zeroes \\= (\alpha + \beta)\Big(\frac{1}{\alpha} + \frac{1}{\beta} \Big)\\= (\alpha + \beta)\Big (\frac{\beta+\alpha}{\alpha \beta }\Big) \\= 2\times \frac{2}{5} \\=\frac{4}{5} \: --(4)$

$\red{ Required \: polynomial : }$

$k [ x^{2} - ( sum \:of \: zeroes)x+product \:of \: zeroes ]$

$= k[ x^{2} - \frac{12}{5}x + \frac{4}{5} ]$

/* For all real values of k it is true. */

$If \: k = 5 ,\: then \\</p><p>required \:polynomial \: 5x^{2} - 12x + 4$

Therefore.,

$\red { Required \: polynomial } \\\green {= 5x^{2} - 12x + 4 }$

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Answer 2

Answer:

5x^2-12x+4

Step-by-step explanation:

alpha+beta=2

alpha.beta=5

sum of zeros=alpha+beta=alpha+beta+1/alpha+1/beta

=>alpha^2.beta+alpha.beta^2+beta+alpha/alpha.beta

=>alpha.beta(alpha+beta)+alpha+beta/alpha.beta

=>5.2+2/5

=>12/5

Product of zeros=alpha.beta=(alpha+beta).(1/alpha+1/beta)

=>1+alpha/beta+beta/alpha+1

=>alpha^2+beta^2/alpha.beta+2

=>(alpha+beta)^2-2.alpha.beta/alpha.beta+2

=>(2)^2-2.5/5+2

=>-6/5+2

=>4/5

required polynomial = x^2-(alpha+beta)x+alpha.beta

=>x^2-12/5+4/5 = 0

=>5x^2-12x+4/5=0

=5x^2-12x+4 Ans.

Thank You