Math, asked by samukchamraxe, 2 months ago

Find the zeroes of the quadratic polynomial 7 2 − 11 3 − 2 3 and verify the relationship between the zeroes and their coefficients.​

Answers

Answered by SKASHISH666
1

Step-by-step explanation:

Multiply the eq by 3

21y2-11y-2=0

21y2-14y+3y-2=0

7y(3y-2)+1(3y-2)=0

(7y+1)(3y-2)=0

y=-1/7 and y = 2/3

a=7 b=-11/3 c=-2/3

sum of roots is -1/7+2/3=11/21 =-b/a

product of root=-2/21=c/a

Answered by 12thpáìn
6

{ \mathbb{QUADRATIC \:  \:  EQUATION } \mapsto \sf7 {x}^{2}  -  \dfrac{11x}{3}  -  \dfrac{2}{3} \:  \:  \:  }

  • Multiplying both sides by 3.

\mapsto \sf  3\bigg(7 {x}^{2}  -  \dfrac{11x}{3}  -  \dfrac{2}{3} \bigg) = 0  \:

\mapsto \sf 21 {x}^{2}  -  11x  -  2 = 0  \:

\mapsto \sf 21 {x}^{2}  -  14x   + 3x-  2 = 0  \:

\mapsto \sf 7x( 3x  - 2)  + 1(3x-  2 )= 0  \:

\mapsto \sf ( 3x  - 2)(7x + 1)= 0  \:

{\mapsto \sf ( 3x  - 2) = 0 \:  \:  \:  \:  \:   \:  \: \:  \:  \:  \:  \:  \:  \: (7x + 1)= 0  \:}

{\mapsto \sf  3x  =2 \:  \:  \:  \:  \:   \:  \: \:  \:  \:  \:  \:  \:  \: 7x= - 1  \:}

{\mapsto \sf  x  = \dfrac{2}{3}  \:  \:  \:  \:  \:   \:  \: \:  \:  \:  \:  \:  \:  \: x=   \dfrac{ - 1}{7}  \:} \\  \\

Let alpha be 2/3 and beta be -1/7.

In Quadratic Equation 7x²-11x/3-2/3=0

On comparing with ax²+bx+c=0

\sf{Sum  \:  \: of \:  \:  zeres (Alpha+beta) = \dfrac{-b}{a}  = \dfrac{-Coefficient  \: of \:  x}{-Coefficient \:  of \:   {x}^{2} }  }

\sf{ \dfrac{2}{3}  -  \dfrac{1}{7}   =  \dfrac{ -(  - \frac{11}{3} )}{7}  }

\sf{ \dfrac{14 - 3}{21}    =  \dfrac{11}{3}   \times  \dfrac{1}{7} }

\sf{ \dfrac{11}{21}    =  \dfrac{11}{21}    }

 \sf{Product \:  of \:  zeros \:  (Alpha×Beta)=\dfrac{c}{a}=\dfrac{ Constant \:  term}{Coefficient \:  of \:  {x}^{2} }}

 \sf{ \dfrac{2}{3} \times  \dfrac{ - 1}{7}  =  \dfrac{ \frac{ - 2}{3}}{7}   }

 \sf{  \dfrac{ - 2}{21}   =  \dfrac{  - 2}{21}   }

Verified

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