A) \[pq{{b}^{2}}+{{(p+q)}^{2}}ac=0\]
B) \[pq{{b}^{2}}-{{(p+q)}^{2}}ac=0\]
C) \[pq{{a}^{2}}-{{(p+q)}^{2}}bc=0\]
D) None of these
Correct Answer: B
Solution :
Let \[p\alpha ,q\alpha \] be the roots of the given equation\[a{{x}^{2}}+bx+c=0\]. Then \[p\alpha +q\alpha =-\frac{b}{a}\]and \[p\alpha .q\alpha =\frac{c}{a}\] From first relation, \[\alpha =-\frac{b}{a(p+q)}\] Substituting this value of \[\alpha \] in second relation, we get \[\frac{{{b}^{2}}}{{{a}^{2}}{{(p+q)}^{2}}}\times pq=\frac{c}{a}\]Þ \[{{b}^{2}}pq-ac{{(p+q)}^{2}}=0\] Note: Students should remember this question as a fact.
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