A) 0
B) 1
C) 2
D) 3
Correct Answer: D
Solution :
Given equations are \[{{x}^{2}}-px+20=0\] \[{{x}^{2}}-20x+p=0\] If \[p=20\] then the quadratic equations are identical. Hence\[x=10+4\sqrt{5}\]or \[x=10-4\sqrt{5}\]. If \[p\ne 20\]then subtracting equations, we get \[(20-p)x+(20-p)=0\] \[\Rightarrow \] \[x=-1\]and\[p=-\,21\] Hence there are 3 values of\[x\].You need to login to perform this action.
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