A) \[{{(ac)}^{1/3}}+{{(ab)}^{1/3}}+c=0\]
B) \[{{({{a}^{3}}b)}^{1/4}}+{{(a{{b}^{3}})}^{1/4}}+c=0\]
C) \[{{({{a}^{3}}c)}^{1/4}}+{{(a{{c}^{3}})}^{1/4}}+b=0\]
D) \[{{({{a}^{4}}c)}^{1/3}}+{{(a{{c}^{4}})}^{1/3}}+b=0\]
Correct Answer: C
Solution :
Since, \[\alpha \] and \[\beta \] are the roots of the equation\[a{{x}^{2}}+bx+c=0\]. \[\therefore \] \[\alpha +\beta =-\frac{b}{a}\]and\[\alpha \beta =\frac{c}{a}\] \[\Rightarrow \]\[\alpha +{{\alpha }^{1/3}}=-\frac{b}{a}\]and\[\alpha \cdot {{\alpha }^{1/3}}=\frac{c}{a}\] \[(\because \,\,given\,\,\beta ={{\alpha }^{1/3}})\] \[\Rightarrow \] \[\alpha ={{\left( \frac{c}{a} \right)}^{3/4}}\] \[\therefore \] \[{{\left( \frac{c}{a} \right)}^{3/4}}+{{\left( \frac{c}{a} \right)}^{1/4}}=-\frac{b}{a}\] \[\Rightarrow \] \[{{({{a}^{3}}c)}^{1/4}}+{{(a{{c}^{3}})}^{1/4}}+b=0\]
You need to login to perform this action.
You will be redirected in
3 sec
