Question

find the zeroes of 3x^2 -10x +7 and also verify the relation between the zeroes​

Answer 1

Step-by-step explanation:

$3 {x}^{2} - 10x + 7 = 0$

$3 {x}^{2} - 3x - 7x + 7 = 0$

$3x(x - 1) - 7(x - 1) = 0$

$(x - 1)(3x - 7) = 0$

$x = 1 \: \: \: \: \: \: or \: \: \: \: \frac{7}{3}$

$Let, \alpha \: and \: \beta \: be \: the \: zeroes \: of \: the \: quadratic \: polynomial.$

$\alpha = 1 \: \: and \: \: \: \beta = \frac{7}{3}$

$\bold{We \: know \: that,}$

$sum \: of \: zeroes = \alpha + \beta = \frac{ - b}{a}$

$\implies \: 1 + \frac{7}{3} = \frac{ - 10}{ 3}$

$\implies \frac{3 + 7}{3} = \frac{10}{3}$

$\implies \frac{10}{3} = \frac{10}{3}$

$product \: o f \: its \: zeroes = \alpha \times \beta = \frac{c}{a}$

$\implies1 \times \frac{7}{3} = \frac{7}{3}$

$\implies \frac{7}{3} = \frac{7}{3}$

Hence verified

Answer 2

Given Polynomial : 3x² -10x +7

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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$\begin{gathered}\qquad:\implies\sf 3x^2 - 10\:x + 7 = 0\\\\\\ \qquad:\implies\sf 3x^2 - 3x - 7x + 7 = 0\\\\\\ \qquad:\implies\sf 3x(x - 1) - 7(x - 1) = 0\\\\\\ \qquad:\implies\sf (x - 1)(3x - 7) = 0\\\\\\ \qquad:\implies{\underline{\boxed{\textsf{\red{x = 1\:\:or\:\:x = 7/3}}}}}\:\bigstar\\\\\\\end{gathered}$

$\therefore\:{\underline{\sf{Hence,\:The\:zeroes\:of\:Polynomial\:are\:{\pmb{1}}\:{\sf{\&}}\:{\bf{7/3}}.}}}$

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

Here, In the given Polynomial 3x² -10x +7, {a = 3 , b = -10 & c = 7}.

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

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

$\begin{gathered}\qquad\quad\dashrightarrow\sf \alpha + \beta = \dfrac{-b}{a}\\\\\\ \qquad\quad\dashrightarrow\sf \big(1\big) + \bigg(\dfrac{7}{3}\bigg) = \dfrac{10}{3}\\\\\\ \qquad\quad\dashrightarrow\sf \dfrac{3}{3} + \dfrac{7}{3} = \dfrac{10}{3}\\\\\\ \qquad\quad\dashrightarrow{\boxed{\boxed{\sf{\pink{\dfrac{10}{3} =\dfrac{10}{3}}}}}}\\\\\\\end{gathered}$

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

$\begin{gathered}\qquad\quad\dashrightarrow\sf \alpha \beta = \dfrac{c}{a}\\\\\\ \qquad\quad\dashrightarrow\sf \big(1\big) \bigg(\dfrac{7}{3}\bigg) = \dfrac{7}{3}\\\\\\ \qquad\quad\dashrightarrow{\boxed{\boxed{\sf{\pink{ \frac{7}{3} = \frac{7}{3} }}}}}\\\\\\\end{gathered}$

Hence Verified !