Let $a0, b0$, and $c0$. Then both the roots of the equation $a x^{2}+b x+c=0$.\

Question

Let $a>0, b>0$, and $c>0$. Then both the roots of the equation $a x^{2}+b x+c=0$.\

Professor PC Thomas Classes · Answer

(b) : $x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$\ $=\frac{-b}{2 a}+\frac{\sqrt{b^{2}-4 a c}}{2 a}, \frac{-b}{2 a}-\frac{\sqrt{b^{2}-4 a c}}{2 a}$\ Case (i): Let $b^{2}-4 a c>0$\ Since $a, b, c>0$, both roots are negative Case (ii) : Let $b^{2}-4 a c=0, x=\frac{-b}{2 a}, \frac{-b}{2 a}$.\ $	herefore \quad$ both roots are negative.\ Case (iii) : Let $b^{2}-4 a c<0$, then\ $x=\frac{-b}{2 a} \pm i \frac{\sqrt{4 a c-b^{2}}}{2 a}$.\ Clearly real part is -ve.\ In all cases both roots have negative real part.\