A) \[p+\frac{1}{p}=1,p\ne 0\]
B) \[{{p}^{2}}+\frac{1}{p}=1,p\ne 0\]
C) \[{{x}^{2}}-\frac{1}{x}=1,x\ne 0\]
D) \[{{x}^{2}}+2\sqrt{x}-1=0\]
Correct Answer: A
Solution :
A quadratic equation has a degree 2. In and, the degree of the polynomial is 3. In, \[\text{1}00\text{1}=\text{91}\times \text{1}0+\text{91}\]is not a polynomial as \[\text{91}0=\text{91}\times \text{1}0+0\]the power of the variable is not an integer. In, \[\therefore \]is a quadratic equation.
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