0 then {ax}^{2}+ {bx}+ {c}=0 will have real roots of opposite sign" Statement: 2 Let us take \alpha and \beta roots of equation a x^{2}+b x+c=0 Product of roots = {c} \/ {a} \Rightarrow \alpha \times \beta= {c} \/ {a} If both roots are positive or negative, we get \Rightarrow {c} \/ {a} > 0 If one root is negative and one is positive, we get \Rightarrow {c} \/ {a}" >
Concept : ax2+bx+c=0 Discriminant (D)=b2−4ac Product of roots =c∕a Calculations : ax2+bx+c=0 Statement: 1D>0 ⇒b2−4ac>0 If a & c both are of opposite sign, ⇒b2−4(−ve)>0 ⇒ Product of root =c∕a<0 This shows both roots should have opposite signs. But the opposite sign of the root will also be possible when both have the same sign. For example, Let us take an example to understand ⇒x2−5x+6=0 ⇒a=1,b=−5,c=6 ⇒D=(−5)2−4×1×6=24 ⇒D>0 Hence, we can not say, "If D>0 then ax2+bx+c=0 will have real roots of opposite sign" Statement: 2 Let us take α and β roots of equation ax2+bx+c=0 Product of roots =c∕a ⇒α×β=c∕a If both roots are positive or negative, we get ⇒c∕a>0 If one root is negative and one is positive, we get ⇒c∕a<0. ∴ Statement − II alone is sufficient to answer the question
