Question

If alpha and beta are the zeroes of x²+x-2 ,form a quadratic polynomial whose zeroes are. 2 alpha +1, 2beta +1
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Answer 1

Answer:

Step-by-step explanation:

x²-3x+2=0

=x²-x-2x+2=0

=x(x-1)-2(x-1)=0

(x-1)(x-2)=0x=1

or ,

x=2 given zeroes of quadratic polynomial 1/2alpha+beta

1/2beta+alpha substitute=1/2(1)+2

1/2(2)+1=1/4

1/5formula is x²-(alpha+beta)x+alpha*beta =0therefore x²-

(1/4+1/5)x+1/4*1/5=0=ans is 20x^2 - 9x +1 =0

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Answer 2

Answer :

$p(x) = {x}^{2} + x - 2 \\ \\ \alpha \: and \: \beta \: are \: the \: roots$

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

$= \frac{ - 1}{1}$

$= - 1$

$\alpha \beta = \frac{c}{a} \\ \\ = \frac{ - 2}{1} \\ \\ = - 2$

$2 \alpha + 1 \: and \: 2 \beta + 1$

are the roots of the equation :

$f(x) = {x}^{2} - (sum \: of \: roots) + product \: of \: roots$

$sum \: of \: roots \: = 2 \alpha + 1 + 2 \beta + 1 \\ = 2 + 2 \alpha + 2 \beta \\ =2 (1 + \alpha + \beta )$

$= 2(1 { - 1}) \\ = 2(0) \\ = 0$

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

$= - 10 + 1 \\ = - 9$

$f(x) = {x}^{2} - 0x - 9 \\ = {x }^{2} - 9$

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