Question

find the quadratic polynomial whose zeros are 2/3 and -6 verify the relationship between coefficients and the zeros of the polynomial

Answer 1

Heya !

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Here given , the zeroes are$\frac{2}{3} and -6 \\ polynomial =\ \textgreater \ x^{2} -sum of zeroes (x)+ product of zeroes \\ =\ \textgreater \ sum of zeroes= 2/3+(-6)=2/3-6 =2-18/3 =-16/3 \\ =\ \textgreater \ product of zeroes=\ \textgreater \ 2/3(-6)=-12/3=-4. \\. put the value in polynomial formula . \\ =\ \textgreater \ x^{2} -(-16/3)x+(-4) \\ =\ \textgreater \ x^{2} +16/3 x -4=o \\ =\ \textgreater \ 3 x^{2} +16x-12 is the required polynomial.. \\ here a= 3 ,b=16 ,c=-12 \\ \alpha + \beta =-b/a=-16/3 \\ \alpha \beta =c/a=-12/3=-4. \\ now verify. \\ =\ \textgreater \ \alpha + \beta =-16/3. \\ =\ \textgreater \ 2/3+(-6)=-16/3. \\ =\ \textgreater \ -16/3=-16/3$

=>@ß= -4
=> 2/3(-6)=-4
=>-4=-4 .

hence verified

hope this helps you ☺

Answer 2

hi very easy question, try it yourself