If a 0 and b 0,what is the nature of the | vedclass

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(D) The given equation is $a - bx - x^2 = 0$,which can be rewritten as $x^2 + bx - a = 0$.Comparing this with the standard quadratic equation $Ax^2 +

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Opposite signs and the numerically larger root is negative

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(D) The given equation is $a - bx - x^2 = 0$,which can be rewritten as $x^2 + bx - a = 0$.Comparing this with the standard quadratic equation $Ax^2 + Bx + C = 0$,we have $A = 1$,$B = b$,and $C = -a$.The discriminant $D$ is given by $D = B^2 - 4AC = b^2 - 4(1)(-a) = b^2 + 4a$.Since $a > 0$ and $b > 0$,$D = b^2 + 4a > 0$,so the roots are real and distinct.Let the roots be $\alpha$ and $\beta$. The product of the roots is $\alpha \beta = \frac{C}{A} = -a$.Since $a > 0$,the product of the roots is negative,which implies the roots have opposite signs.The sum of the roots is $\alpha + \beta = -\frac{B}{A} = -b$.Since $b > 0$,the sum of the roots is negative. For the sum of two numbers with opposite signs to be negative,the numerically larger root must be negative.

If $a > 0$ and $b > 0$,what is the nature of the roots of the equation $a - bx - x^2 = 0$?

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