Question

Find the zeros of following quadratic polynomial and verify the relationship between the zeros and co-efficient of X^-5X+6

Answer 1

Given Polynomial : x² - 5x + 6

We've to find relationship b/w zeroes and Coefficient of given quadratic Polynomial.

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☆ Let's find out zeroes of Given Polynomial :

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$\begin{gathered}\qquad:\implies\sf x^2 - 5x + 6= 0\\\\\\ \qquad:\implies\sf x^2 - 2x - 3x+6 = 0\\\\\\ \qquad:\implies\sf x(x - 2) - 3(x - 2)= 0\\\\\\ \qquad:\implies\sf (x - 2)(x - 3) = 0\\\\\\ \qquad:\implies{\underline{\boxed{\pmb{\frak{\red{x = 2\:\:or\:\:x = 3}}}}}}\:\bigstar\\\\\\\end{gathered}$

$\therefore\:{\underline{\sf{Hence,\:The\:zeroes\:of\:Polynomial\:are\:{\pmb{ \sf{ 2}}}\:{\sf{\&}}\:{\bf{3}}.}}}$

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✇ Let's consider α and β be zeroes of Polynomial.

Here, In the given Polynomial x² - 5x + 6, {a = 1 , b = 5 & c = 6}.

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☆ Now, Let's verify the relationship between zeroes and Coefficient :

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$\begin{gathered}\begin{gathered}\bf{\dag}\:{\underline{\boxed{\pmb{\sf{Sum\:of\:zeroes\:\purple{(\alpha + \beta)}\::}}}}}\\\\\\\end{gathered} \end{gathered}$

$\begin{gathered}\begin{gathered}\qquad\quad\dashrightarrow\sf \alpha + \beta = \dfrac{-b}{a}\\\\\\ \qquad\quad\dashrightarrow\sf \bigg( + 2\bigg) + \bigg( + 3\bigg) = \dfrac{5}{1}\\\\\\ \qquad\quad\dashrightarrow\sf + 2 + 3= 5\\\\\\ \qquad\quad\dashrightarrow{\boxed{\boxed{\frak{\pink{ 5 = 5}}}}}\\\\\\\end{gathered} \end{gathered}$

$\begin{gathered}\begin{gathered}\bf{\dag}\:{\underline{\boxed{\pmb{\sf{Product\:of\:zeroes\:\purple{(\alpha \beta)}\::}}}}}\\\\\\\end{gathered} \end{gathered}$

$\begin{gathered}\begin{gathered}\qquad\quad\dashrightarrow\sf \alpha \beta = \dfrac{c}{a}\\\\\\ \qquad\quad\dashrightarrow\sf \bigg( + 2\bigg) \bigg( + 3\bigg) = \dfrac{6}{1}\\\\\\ \qquad\quad\dashrightarrow\sf 3 \times 2 = 6\\\\\\ \qquad\quad\dashrightarrow{\boxed{\boxed{\frak{\pink{6= 6}}}}}\\\\\\\end{gathered} \end{gathered}$

Hence Verified !