Question

find the quadratic polynomial sum of whose zeroes is 2underoot 3 and their product is 2?​

Answer 1

$Let the polynomial be \ p(x)=0. \\ \\ Let the zeroes of the polynomial be \ \alpha\ \&\ \beta.$

$\alpha+\beta=2\sqrt{3} \ \ \ \ \ \longrightarrow \ \ \ \ \ (1) \\ \\ Let \ \ m=2\sqrt{3} \\ \\ \\ \alpha\beta=2 \\ \\ Let \ \ n=2$

$\therefore\ \alpha-\beta=\sqrt{m^2-4n} \\ \\ \Rightarrow \alpha-\beta=\sqrt{(2\sqrt{3})^2-(4\times 2)} \\ \\ \Rightarrow \alpha-\beta=\sqrt{12-8} \\ \\ \Rightarrow \alpha-\beta=\sqrt{4} \\ \\ \alpha-\beta=2 \ \ \ \ \ \longrightarrow \ \ \ \ \ (2)$

$(1)+(2) \\ \\ (\alpha+\beta)+(\alpha-\beta)=2\sqrt{3}+2 \\ \\ 2\alpha=2\sqrt{3}+2 \\ \\ \alpha=\frac{2\sqrt{3}+2}{2} \\ \\ \alpha=\sqrt{3}+1$

$(1)-(2) \\ \\ (\alpha+\beta)-(\alpha-\beta)=2\sqrt{3}-2 \\ \\ 2\beta=2\sqrt{3}-2 \\ \\ \beta=\frac{2\sqrt{3}-2}{2} \\ \\ \beta=\sqrt{3}-1$

$So let's write \ p(x). \\ \\ \\ p(x)=(x-\alpha)(x-\beta) \\ \\ p(x)=(x-(\sqrt{3}+1))(x-(\sqrt{3}-1)) \\ \\ p(x)=x^2-(\sqrt{3}+1+\sqrt{3}-1)x+(\sqrt{3}+1)(\sqrt{3}-1) \\ \\ p(x)=\bold{x^2-2\sqrt{3}\ x+2}$

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$But there's no need to use the above method! \\ \\ Just do the following:$

$Let the polynomial be \ p(x)=ax^2+bx+c=0. \\ \\ Let the roots be \ \alpha\ \&\ \beta.$

$Only assume that coefficient of \ x^2,\ i.e., \ a,\ is 1. \\ \\ So we get that, \\ \\ \alpha+\beta=-\frac{b}{a}=\frac{-b}{1}=-b=2\sqrt{3}\ \ \ ; \ \ \ b=-2\sqrt{3} \\ \\ and \\ \\ \alpha\beta=\frac{c}{a}=\frac{c}{1}=c=2 \\ \\ \\ \therefore\ We get easily that \ p(x)=\bold{x^2-2\sqrt{3}\ x+2=0}$

$When such questions are asked, apply the negative value of the given sum as the coefficient of \ x,\ i.e., \ b,\ and the value of the given product as the coefficient of \ x^0,\ i.e., \ c,\ by assuming the value of coefficient of \ x^2,\ i.e., \ a,\ as 1. \\ \\ \\ That's all!!!$

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$Thank you. Have a nice day. :-) \\ \\ \\ \\ \\ \#adithyasajeevan$