Question

find the zeroes of polynomial 3x square -7x +4 and verify the relation between zeros and cooficients​

Answer 1

$\huge\purple{\underline{\underline{\pink{Ans}\red{wer:-}}}}$

$\sf{Zeroes \ of \ polynomial \ are \ 1 \ and \ \frac{4}{3}}$

$\sf\orange{Given:}$

$\sf{The \ given \ polynomial \ is}$

$\sf{\implies{3x^{2}-7x+4}}$

$\sf\pink{To \ find:}$

$\sf{Zeroes \ of \ polynomial.}$

$\sf\green{\underline{\underline{Solution:}}}$

$\sf{The \ given \ polynomial \ is}$

$\sf{\implies{3x^{2}-7x+4}}$

$\sf{\implies{3x^{2}-3x-4x+4}}$

$\sf{\implies{3x(x-1)-4(x-1)}}$

$\sf{\implies{(x-1)(3x-4)}}$

$\sf{\implies{x=1 \ or \ \frac{4}{3}}}$

$\sf\purple{\tt{\therefore{Zeroes \ of \ polynomial \ are \ 1 \ and \ \frac{4}{3}}}}$

$\sf\blue{Verification:}$

$\sf{The \ given \ polynomial \ is}$

$\sf{\implies{3x^{2}-7x+4}}$

$\sf{Here, \ a=3, \ b=-7 and \ c=4}$

$\sf{Let \ \alpha \ be \ 1 \ and \ \beta \ be \ \frac{4}{3}}$

________________________________

$\sf{Sum \ of \ zeroes=\frac{-b}{a}}$

$\sf{\implies{\alpha+\beta=1+\frac{4}{3}}}$

$\sf{\implies{\alpha+\beta=\frac{3+4}{3}}}$

$\sf{\implies{\alpha+\beta=\frac{7}{3}...(1)}}$

$\sf{\implies{\frac{-b}{a}=\frac{-(-7)}{3}}}$

$\sf{\implies{\frac{-b}{a}=\frac{7}{3}...(2)}}$

$\sf{...from \ (1) \ and \ (2)}$

$\sf{\implies{\alpha+\beta=\frac{-b}{a}}}$

________________________________

$\sf{Product \ of \ zeroes=\frac{c}{a}}$

$\sf{\implies{\alpha\beta=1(\frac{4}{3})}}$

$\sf{\implies{\alpha\beta=\frac{4}{3}...(3)}}$

$\sf{\implies{\frac{c}{a}=\frac{4}{3}...(4)}}$

$\sf{...from \ (3) \ and \ (4)}$

$\sf{\implies{\alpha\beta=\frac{c}{a}}}$

$\sf{Hence, \ verified.}$

Answer 2

$\bold{\red{\underline{\underline{\green{Solution:}}}}}$

$\tt{i) \:p(x) = {3x}^{2} - 7x + 4}$

Now factorise it first

3x² - 7x + 4

$\implies$ 3x² + (4 + 3) x + 4

$\implies$ 3x² - 4x - 3x + 4

$\implies$ x (3x - 4) - 1 (3x - 4)

$\implies$ (x - 1) (3x - 4)

Now, x - 1 = 0 $\implies$ x = 1

3x = 4 = 0 $\implies$ x = $\frac{4}{3}$

Now, 1 and $\large{\frac{4}{3}}$ are the two root of the polynomial

Let, 1 = $\alpha$ + $\beta$ = $\Large{\frac{-b}{a}}$

$\implies$ 1 + $\large{\frac{4}{3}}$ = $\large{\frac{-(-7)}{3}}$

$\implies$ $\large{\frac{3+4}{3}}$ = $\large{\frac{7}{3}}$

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

ii) $\alpha$ $\beta =\large{ \frac{c}{a}}$

$1×\large{\frac{4}{3}} = \large{\frac{4}{3}}$

$\implies$ $\large{\frac{4}{3}}$ $=\large{\frac{4}{3}}$ Proved