A) \[{{x}^{2}}-2x+4=0\]
B) \[{{x}^{2}}-2x+16=0\]
C) \[{{x}^{2}}-4x+8=0\]
D) \[{{x}^{2}}-4x+16=0\]
Correct Answer: D
Solution :
Let \[\alpha ,\beta \]be roots of \[{{x}^{2}}\]- 2x + 4 = 0 Then, \[\alpha +\beta =-\left( -2 \right)=2\Rightarrow \alpha \beta =4\] Quadratic equation whose roots are\[2\alpha ,\,\,2\beta \]is \[{{x}^{2}}-\left( 2\alpha +2\beta \right)x+2\alpha \times 2\beta =0\] \[\Rightarrow \,{{x}^{2}}-2\left( \alpha +\beta \right)x+4\alpha \beta =0\] \[\]
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