Math, asked by ashikaushik7612, 11 months ago

If alfha and beta are the zeroes of the polynomial,such that alpha+beta=6 and alpha×beta=4,write the polynomial

Answers

Answered by shadowsabers03
1

Consider a polynomial,...

p(x)=ax^2+bx+c=0

If  α  and  β  are the zeroes of p(x), then,

\begin{aligned}1.&\ \ \alpha+\beta&=&\ \ -\dfrac{b}{a}\\ \\ \\ 2.&\ \ \alpha\beta&=&\ \ \dfrac{c}{a}\end{aligned}

Given that,

\alpha+\beta=6\ \ \ \ \ \&\ \ \ \ \ \alpha\beta=4

Let the coefficient of  x²  of p(x), i.e.,  a  be 1.

\textsf{Let}\ \ \ a=1.\\ \\ \\ \begin{aligned}&\alpha+\beta&=&\ \ 6\\ \\ \Longrightarrow\ \ &-\frac{b}{a}&=&\ \ 6\\ \\ \Longrightarrow\ \ &-\frac{b}{1}&=&\ \ 6\\ \\ \Longrightarrow\ \ &-b&=&\ \ 6\\ \\ \Longrightarrow\ \ &\ \ \ \ b&=&\ \ -6\\ \\ \\ &\alpha\beta&=&\ \ 4\\ \\ \Longrightarrow\ \ &\frac{c}{a}&=&\ \ 4\\ \\ \Longrightarrow\ \ &\frac{c}{1}&=&\ \ 4\\ \\ \Longrightarrow\ \ &c&=&\ \ 4\end{aligned}

Thus,

\large \text{$a=1\ \ \ ; \ \ \ b=-6\ \ \ ; \ \ \ c=4$}

Now, we can make a polynomial.

\Large \text{$\tt{p(x)=x^2-6x+4=0}$}

Hence a polynomial is found! We can make more other polynomials by giving many other values for  a.

\textsf{Let}\ \ a=2. \\ \\ \\ \begin{aligned}&\alpha+\beta&=&\ \ 6\\ \\ \Longrightarrow\ \ &-\frac{b}{a}&=&\ \ 6\\ \\ \Longrightarrow\ \ &-\dfrac{b}{2}&=&\ \ 6\\ \\ \Longrightarrow\ \ &\ \ \ \ b&=&\ \ -12\\ \\ \\ &\alpha\beta&=&\ \ 4\\ \\ \Longrightarrow\ \ &\frac{c}{a}&=&\ \ 4\\ \\ \Longrightarrow\ \ &\frac{c}{2}&=&\ \ 4\\ \\ \Longrightarrow\ \ &c&=&\ \ 8\end{aligned}

Here another polynomial is formed.

\large \text{$a=2\ \ \ ; \ \ \ b=-12\ \ \ ; \ \ \ c=8$}\\ \\ \\ \\ \Large \text{$\tt{p(x)=2x^2-12x+8=0}$}

Here we can find out that the polynomial obtained here is  'a'  multipled by the first polynomial.

2x^2-12x+8=2(x^2-6x+4)\ \ \ \ \ \ \ \ \ \ [a=2]

Hence we can say one thing.

\begin{flushleft}\sf Any second degree polynomial in the form \ $n(x^2-6x+4)$ \ or \ $nx^2-6nx+4n$ \ having zeroes \ $\alpha$ \ and \ $\beta$ \ always satisfy the following:\end{flushleft}\\ \\ \\ \begin{center}\large \text{$\alpha+\beta=6$}\\ \\ \\ \large \text{$\alpha\beta=4$}\end{center}\\ \\ \begin{flushleft}\normalsize for any real number $n$.\end{flushleft}

Here,  a = n.

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