Question

If alpha and beta are the zeros of the quadratic polynomial x^2+x-2 find a polynomial whose zeros are 2alpha+1 and 2Beta-1

Answer 1

Hey!

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x^2 + x - 2

a = 1
b = 1
c = -2

Sum of zeroes ( @ + ß ) = - b/a = -1/1
Product of zeroes ( @ß) = c/a = -2/1

A.T.Q

We have to find a polynomial whose zeroes = 2@+1 and 2ß+1

Let find sum and product of zeroes -

Sum of zeroes = 2@+1 + ( 2ß-1)

= 2@ + 2ß
= 2 ( @ + ß)
= 2 × -1/1
= -2


Product of zeroes = (2@ + 1)(2ß-1)
= 4@ß - 2@ + 2ß - 1
= 4 × -2/1 - 2 × (1) - 1
= -8 - 2 - 1
= -11


Quadratic polynomial =

k ( x^2 - (@ + ß) x + @ß)
= x^2 - ( -2) x + (-11)
= x^2 + 2 - 11


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Hope it helps...!!!

Answer 2

$Hey \: There \:$ !!

$p(x) = {x}^{2} + x - 2$

$Given \: that,$ $\alpha \: and \: \beta \: are \: the \: zeroes \: of \: given \: polynomial.$

$sum \: of \: zeroes = \alpha + \beta$
$= \frac{ - b}{a} = \frac{ - 1}{1} = 1$

$product \: of \: zeroes = \alpha \beta$
$= \frac{c}{a} = \frac{ - 2}{1} = - 2$

$Now \: A.T.Q.$,
$sum \: of \: zeroes = (2 \alpha + 1) + (2 \beta - 1)$
$= 2\alpha + 2 \beta$
$= 2( \alpha + \beta )$
$= 2( - 1)$
$= - 2$

$product \: of \: zeroes = (2 \alpha + 1)(2 \beta - 1)$
$= 2 \alpha (2 \beta - 1) + (2 \beta - 1)$
$= 4 \alpha \beta - 2 \alpha + 2 \beta - 1$
$= 4( - 2) - 2(1) - 1$
$= - 8 - 2 - 1$
$= - 11$

$We \: know \: that,$
$Quadratic \: polynomial = {x}^{2} - ( \alpha + \beta )x + \alpha \beta$
$= {x}^{2} + 2x - 11$

$Hope \: my \: ans.'s \: satisfactory$ !!