Q. If the sum of square of the zeros of the polynomials 6x^2+x+k is 25/36 then find the value of K ?
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We know that in ax^2+bx+c,
sum of zeroes is -b/a
product of zeroes= c/a
so, in this polynomial,
sum of zeroes= -1/6
product of zeroes= p/6
by using (a+b)^2= a^2+b^2+2ab
(-1/6)^2= 25/36+2p/6 {if a & b are zeros}
1/36= 25/36+ 12p/36
=1= 25+12p
=12p= 1-25
=p= -24/12
therefore= -2 Ans.
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