Question

plz solve fastttttttt ⚡⚡​

Answer 1

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

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

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

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

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

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

Answer 2

Question :-

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

• 1/4 , -1

• 1,1

• 4,1

Solution :-

(1.) 1/4 , -1

Sum of zeros = $( \alpha + \beta ) = 1/4$

Product of zeros = $\alpha \beta = -1$

As we know that quadratic polynomial is,

$\large\implies \bf {ax}^{2} + bx + c$

$\\ \large \implies \bf k [ {x}^{2} - ( \alpha + \beta )x + \alpha \beta ]$

Where K is an constant term and $k \cancel = 0$,then

$\large \implies \bf k[ {x}^{2} - ( \frac{1}{4} )x + ( - 1)]$

$\\ \large \implies \bf {4x}^{2} - x - 1$

(2). 1,1

Sum of zeros = $( \alpha + \beta ) = 1$

Product of zeros = $\alpha \beta = 1$

$\large \implies \bf k[{x}^{2} - (1)x + (1)]$

$\\ \large \implies \bf {x}^{2} - x + 1$

(3). 4,1

Sum of zeros = $( \alpha + \beta ) = 4$

Product of zeros = $\alpha \beta = 1$

$\large \implies \bf k[{x}^{2} - (4)x + (1)]$

$\\ \large \implies \bf {4x}^{2} - x + 1$