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Maths-

General

Easy

Question

If $blank alpha text end text$ and $text end text beta$ are the roots of the equation $blank a x to the power of 2 end exponent plus b x plus c equals 0$ then the quadratic equation whose roots are $blank fraction numerator alpha over denominator 1 plus alpha end fraction text and end text fraction numerator beta over denominator 1 plus beta end fraction text is end text$

$a x^{2} - b (1 - x) + c (1 - x)^{2} = 0$
$a x^{2} - b x (x - 1) + c (x - 1)^{2} = 0$
$a x^{2} + b (1 - x) + c (1 - x)^{2} = 0$
$a x^{2} + b (x + 1) + c (1 + x)^{2} = 0$

The correct answer is: $a x to the power of 2 end exponent plus b left parenthesis 1 minus x right parenthesis plus c left parenthesis 1 minus x right parenthesis to the power of 2 end exponent equals 0$

$text Since roots of end text a x to the power of 2 end exponent plus b x plus c equals 0 text are end text alpha text and end text beta text , hence, roots of end text$
$c x to the power of 2 end exponent plus b x plus a equals 0$
$text will be end text fraction numerator 1 over denominator alpha end fraction text and end text fraction numerator 1 over denominator beta end fraction text . Now replacing end text x rightwards arrow x minus 1 comma text then roots of end text$
$c left parenthesis x minus 1 right parenthesis to the power of 2 end exponent plus b left parenthesis x minus 1 right parenthesis plus a equals 0$
$text will be end text 1 plus fraction numerator 1 over denominator alpha end fraction text and end text 1 plus fraction numerator 1 over denominator beta end fraction text . Now replacing end text x rightwards arrow fraction numerator 1 over denominator x end fraction text we will ge end text text t end text$
$c left parenthesis 1 minus x right parenthesis to the power of 2 end exponent plus b left parenthesis 1 minus x right parenthesis plus a x to the power of 2 end exponent equals 0$
$text whose roots are end text fraction numerator alpha over denominator 1 plus alpha end fraction text and end text fraction numerator beta over denominator 1 plus beta end fraction text . end text$

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