Question

Find the zeroes of the quadratic polynomial 6x^2+x-2 and verify the relationship between the zeroes and the coefficients? ​

Answer 1

Answer:

$\sf{The \ zeroes \ of \ the \ polynomial} \\ \sf{are \ \dfrac{-2}{3} \ and \ \dfrac{1}{2}.}$

Given:

$\sf{The \ given \ quadratic \ polynomial \ is} \\ \sf{6x^{2}+x-2}$

To find:

$\sf{The \ zeroes \ of \ the \ polynomial.}$

Solution:

$\sf{f(x)=6x^{2}+x-2; \ f(x)=0} \\ \\ \sf{6x^{2}+x-2=0} \\ \\ \sf{\therefore{6x^{2}+4x-3x-2=0}} \\ \\ \sf{\therefore{2x(3x+2)-(3x+2)=0}} \\ \\ \sf{\therefore{(3x+2)(2x-1)=0}} \\ \\ \sf{\therefore{x=\dfrac{-2}{3} \ and \ \dfrac{1}{2}}} \\ \\ \purple{\tt{\therefore{The \ zeroes \ of \ the \ polynomial \ are \ \dfrac{-2}{3} \ and \ \dfrac{1}{2}.}}}$

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Verification:

$\sf{The \ given \ polynomial \ is} \\ \\ \sf{f(x)=6x^{2}+x-2} \\ \\ \sf{Here, \ a=6, \ b=1 \ and \ c=-2} \\ \\ \sf{Let \ \alpha=\dfrac{-2}{3} \ and \ \beta=\dfrac{1}{2}} \\ \\ \sf{\alpha+\beta=\dfrac{-2}{3}+\dfrac{1}{2}} \\ \\ \sf{\therefore{\alpha+\beta=\dfrac{-1}{6}...(1)}} \\ \\ \sf{\dfrac{-b}{a}=\dfrac{-1}{6}...(2)} \\ \\ \sf{From \ (1) \ and \ (2), \ we \ get} \\ \\ \boxed{\sf{Sum \ of \ zeroes=\dfrac{-b}{a}}} \\ \\ \sf{\alpha\beta=\dfrac{-2}{3}\times\dfrac{1}{2}} \\ \\ \sf{\therefore{\alpha\beta=\dfrac{-1}{3}...(3)}} \\ \\ \sf{\dfrac{c}{a}=\dfrac{-2}{6}=\dfrac{-1}{3}...(4)} \\ \\ \sf{From \ (3) \ and \ (4), \ we \ get} \\ \\ \boxed{\sf{Product \ of \ zeroes=\dfrac{b}{a}}} \\ \\ \sf{Hence, \ verified.}$

Answer 2

Answer:

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Step-by-step explanation:

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