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Question

If b $$\in \beta^+ $$ the roots of the equation $$(2 + b)x^2 + (3 + b )x + (4 + b) = 0 $$ is 


Solution

$$ (2+b)x^{2}+(3+b)x+(4+b) = 0 $$
$$ x = \dfrac{-(3+b)\pm \sqrt{(3+b)^{2}-4(4+b)(2+b)}}{2(2+b)} $$
$$ = \dfrac{-(3+b)\pm \sqrt{9+b^{2}+6b-4(b^{2}+6b+8)}}{2(2+b)} $$
$$ = \dfrac{-(3+b)\pm \sqrt{9+b^{2}+6b-4b^{2}-24b-32}}{2(2+b)} $$
$$ = \dfrac{-(3+b)\pm \sqrt{-3b^{2}-18b-2}}{2(2+b)} $$
it has imaginary Roots 
They will be the Roots of the 
$$ Eq^{n} $$.

1097505_1138203_ans_4d61a5afafe241e195ec168d341cc09e.jpg

Mathematics

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