If alpha and beta be the zeroes of polynomial x²+5x+k where (alpha) ²+(beta)²=11 find k
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Zeroes= a and b
P(x)=x^2+5x+k
a^2+b^2=11
Now,
Sum of zeroes=a+b=>-b/a=>5/1
Product of zeroes=ab=>c/a=>k/1
Now,
(a+b)^2=a^2+b^2+2ab
(5)^2=11+2k
25=11+2k
25-11=2k
14=2k
14/2=k
Therefore,k=7
P(x)=x^2+5x+k
a^2+b^2=11
Now,
Sum of zeroes=a+b=>-b/a=>5/1
Product of zeroes=ab=>c/a=>k/1
Now,
(a+b)^2=a^2+b^2+2ab
(5)^2=11+2k
25=11+2k
25-11=2k
14=2k
14/2=k
Therefore,k=7
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