Question

Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively

Answer 1

x^2-(sum of zeros)x+product of zeros

Answer 2

Step-by-step explanation:

$\underline{\:\large{\textit{1) \ \ \sf 1/4, -1 \: :}}}$

$\frak{Given}\begin{cases}\sf{Sum \ of \ Zeroes \: (\alpha \: + \: \beta) = \frac{1}{4}}\\\sf{Product \ of \ Zeroes = \ (\alpha \ \beta) = -1}\end{cases}$

$\underline{\bigstar\:\textsf{Required Quadratic Polynomial :}}$

$:\implies\sf p(x) = x^2 - (\alpha + \beta)x + \alpha \: \beta \\\\\\:\implies\sf p(x) = x^2 - \dfrac{1}{4}x + (- 1) \\\\\\:\implies\sf p(x) = x^2 - \dfrac{1}{4}x - 1 \\\\\\:\implies\boxed{\frak{\purple{p(x) = 4x^2 - x -1}}}$

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$\underline{\:\large{\textit{2) \ \ \sf 1, 1 \: :}}}$

$\frak{Given}\begin{cases}\sf{Sum \ of \ Zeroes \: (\alpha \: + \: \beta) = 1 }\\\sf{Product \ of \ Zeroes \: (\alpha \ \beta) = 1}\end{cases}$

$\underline{\bigstar\:\textsf{Required Quadratic Polynomial :}}$

$:\implies\sf p(x) = x^2 - (\alpha + \beta)x + \alpha \: \beta \\\\\\:\implies\sf p(x) = x^2 - 1x + 1 \\\\\\:\implies\boxed{\frak{\purple{p(x) = x^2 - x + 1}}}$

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$\underline{\:\large{\textit{3) \ \ \sf 4,1 \: :}}}$

$\frak{Given}\begin{cases}\sf{Sum \ of \ Zeroes \: (\alpha \: + \: \beta) = 4 }\\\sf{Product \ of \ Zeroes \: (\alpha \ \beta) = 1}\end{cases}$

$\underline{\bigstar\:\textsf{Required Quadratic Polynomial :}}$

$:\implies\sf p(x) = x^2 - (\alpha + \beta)x + \alpha \: \beta \\\\\\:\implies\boxed{\frak{\purple{p(x) = x^2 - 4x + 1}}}$

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$\qquad{\underline{\underline{\bigstar \: \bf \ Some \ Information \ about \ polynomial\: :}}}$⠀

$\boxed{\begin{minipage}{5.5 cm} { \bigstar\: \textsf{For a Quadratic Polynomial :}}\\\\ {\qquad\sf p(x) = ax ^\sf2 \sf + bx + c}\\\sf with zeroes \alpha\:\sf and\:\beta \\\\\\ {\textcircled{\footnotesize1}} \:\:\alpha +\beta= \dfrac{ - \:b}{a}\:\:\bigg\lgroup\bf Sum\:of\:Zeroes\bigg\rgroup \\\\\\{\textcircled{\footnotesize2}} \: \:\alpha \beta= \sf\dfrac{c}{a}\:\:\bigg\lgroup\bf Product\:of\:Zeroes\bigg\rgroup\end{minipage}}$⠀