Question

$whose \: zeroes \: are \: (2 \alpha + 3 \beta ) \: and \: (3 \alpha + 2 \beta )$
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Answer 1

Answer:

Polynomial = x^2 - (sum of zeroes)x + (product)

=》 Sum of zeroes = $\tt{5\alpha + 5\beta}$

=》 Product of zeroes = $\tt{6\alpha^{2} + 13\alpha\beta + 6\beta^{2}}$

Polynomial = $\tt{x^{2} -( 5\alpha + 5\beta)x + 6\alpha^{2} + 13\alpha\beta + 6\beta^{2}}$

Answer 2

Zeroes a and b of the polynomial are

$2 \alpha + 3 \beta \: and \: 3 \alpha + 2 \beta$

So a+ b is equal to

$5 \alpha + 5 \beta$

And ab is equal to

$(2 \alpha + 3 \beta ) \times (3 \alpha + 2 \beta )$

Hence ab =>

$6 \alpha^{2} + 6 \beta^{2} + 12 \alpha \beta$

So the equation is

$x ^{2} - (5 \alpha + 5 \beta )x + (6 \alpha^{2} + 6 \beta ^{2} + 12 \alpha \beta )$

Hope it helps you.