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Question 7
The zeroes of the quadratic polynomial x2+99x+127 are
(A) both positive
(B) both negative
(C) one positive and one negative
(D) both equal

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Solution

The given quadratic polynomial is p(x)=x2+99x+127
On comparing p(x) with ax2+bx+c, we get
a = 1, b = 99 and c = 127
We known that, x=b±b24ac2a
= 99±(99)24×1×1272×1
= 99±98015082
= 99±92932=99±9642
= 99+96.42,9996.42
= 2.62,195.42
= 13,977
Hence, both zeroes of the given quadratic polynomial p (x) are negative
Alternative Method
We know that,
In quadratic polynomial, if a>0a<0 or b>0,c>0b<0,C<0} , then both zeroes are negative
In given polynomial, we see that
a=1>0,b=99>and c=127>0
which satisfy the above condition
so, both zeroes of the given quadratic polynomial are negative

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