A) \[ac{{x}^{2}}+(a+c)bx+{{(a+c)}^{2}}=0\]
B) \[ab{{x}^{2}}+(a+c)bx+{{(a+c)}^{2}}=0\]
C) \[ac{{x}^{2}}+(a+b)cx+{{(a+c)}^{2}}=0\]
D) None of these
Correct Answer: A
Solution :
Here \[\alpha +\beta =-\frac{b}{a}\]and \[\alpha \beta =\frac{c}{a}\] If roots are \[\alpha +\frac{1}{\beta },\beta +\frac{1}{\alpha },\]then sum of roots are \[=\left( \alpha +\frac{1}{\beta } \right)+\left( \beta +\frac{1}{\alpha } \right)=(\alpha +\beta )+\frac{\alpha +\beta }{\alpha \beta }=-\frac{b}{ac}(a+c)\] and product \[=\left( \alpha +\frac{1}{\beta } \right)\text{ }\left( \beta +\frac{1}{\alpha } \right)\] \[=\alpha \beta +1+1+\frac{1}{\alpha \beta }=2+\frac{c}{a}+\frac{a}{c}\] \[=\frac{2ac+{{c}^{2}}+{{a}^{2}}}{ac}=\frac{{{(a+c)}^{2}}}{ac}\] Hence required equation is given by \[{{x}^{2}}+\frac{b}{ac}(a+c)x+\frac{{{(a+c)}^{2}}}{ac}=0\] Þ \[ac{{x}^{2}}+(a+c)bx+{{(a+c)}^{2}}=0\]. Trick: Let\[a=1\], \[b=-3,c=2\], then \[\alpha =1,\]\[\beta =2\] \[\therefore \,\,\,\,\,\,\alpha +\frac{1}{\beta }=\frac{3}{2}\] and \[\beta +\frac{1}{\alpha }=3\] Therefore, required equation must be \[(x-3)(2x-3)=0\]i.e. \[2{{x}^{2}}-9x+9=0\] Here (a) gives this equation on putting \[a=1,b=-3,c=2\].
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