lf $a>0,b>0$, then roots of the equation $a-bx-x^{2}=0$ are:
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both positive
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both negative
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of opposite sign and numerically greater root is positive
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of opposite sign and numerically greater root is negative
The equation is -x^2 - bx + a = 0, or x^2 + bx - a = 0. The product of roots is -a. Since a > 0, the product is negative, meaning the roots have opposite signs. The sum of roots is -b. Since b > 0, the sum is negative, meaning the root with the larger absolute value must be negative.
The given equation is a - bx - x^2 = 0, which can be rewritten in standard quadratic form as x^2 + bx - a = 0. Using the sign rule for roots, the product of the roots is c/a, which here equals -a. Since a is strictly greater than 0, the product of the roots is negative, meaning the roots are of opposite sign. The sum of the roots is -b/a, and since both a and b are greater than 0, the sum of the roots is negative. Having a negative sum with roots of opposite sign guarantees that the numerically larger root must be the negative one.