A) 6
B) 3
C) 2
D) 4
Correct Answer: A
Solution :
Given, \[y_{2}^{3/2}=y_{1}^{1/2}+4\] Squaring on both sides, we get \[y_{2}^{3}={{y}_{1}}+16+8y_{1}^{1/2}\] \[\Rightarrow \] \[{{(y_{2}^{3}-{{y}_{1}}-16)}^{2}}=64{{y}_{1}}\] \[\Rightarrow \] \[y_{2}^{6}-32y_{2}^{3}-2y_{2}^{3}.{{y}_{1}}+y_{1}^{2}-32{{y}_{1}}+256=0\] Hence, the degree of the given equation is 6.You need to login to perform this action.
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