A) 2
B) -2
C) 0
D) 4
Correct Answer: B
Solution :
The discriminant of the quadratic equation\[2{{x}^{2}}+6x+b=0\]is given by\[D=36-8b>0\]. Therefore, the given equation has real roots. we have, \[\frac{{{x}_{1}}}{{{x}_{2}}}=\frac{{{x}_{2}}}{{{x}_{1}}}=\frac{x_{1}^{2}+x_{2}^{2}}{{{x}_{1}}\cdot {{x}_{2}}}=\frac{{{({{x}_{1}}+{{x}_{2}})}^{2}}-2{{x}_{1}}{{x}_{2}}}{{{x}_{1}}\cdot {{x}_{2}}}\] \[=\frac{{{(-3)}^{2}}-2(b/2)}{(b/2)}=\frac{18}{b}-2<-2\] \[[\because \,\,b<0]\]You need to login to perform this action.
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