The correct option is option (b) both negative
$\alpha$ and $\beta$ be the zeroes of $\mathrm{x}^{2}+88 \mathrm{x}+125$
Then $\alpha+\beta=-88$ and $\alpha \times \beta=125$
This can only happen when both the zeroes are negative.
The zeroes of the quadratic polynomial $x^{2}+88 x+125$ are (a) both positive (b) both negative (c) one positive and one negative (d) both equal
The zeroes of the quadratic polynomial $x^{2}+88 x+125$ are (a) both positive (b) both negative (c) one positive and one negative (d) both equal