Question

14 y²-22/3y -4/3 find zeros and verify the relationship between zeroes and coefficients

Answer 1

Answer:

your answer

Step-by-step explanation:

Let f(y)=7y2−113y−23

=21y2−11y−2

=21y2−14y+3y−2 [by splitting the middle term]

=7y(3y−2)+1(3y−2)

=(3y−2)(7y+1)

So, the value of 7y2−113y−23 is zero when 3y−2=0 or 7y+1=0.

i.e., when y=23 or y=−17.

So, the zeroes of 7y2−113y−23 are 23 and −17

∴ Sum of zeroes =23−17=14−321=1121=−(−113×7)

=(−1)(Coefficient of y)(Coefficient ofy2)

and product of zeroes =(23)(−17)=−221=−23×7

=(−1)2(Constant termCoefficient of y2)

Hence, verified the relations between the zeroes and the coefficients of the polynomial.

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