A) \[{{x}^{2}}+9\]
B) \[{{x}^{2}}-4\]
C) \[{{x}^{2}}-9\]
D) \[{{x}^{2}}+4\]
Correct Answer: C
Solution :
Given quadratic polynomial |
\[f\left( x \right)={{x}^{2}}+x-2\] |
\[\alpha +\beta =-1\]and \[\alpha \,.\,\beta =-2\] |
Sum of zeroes i.e., \[\left( 2\alpha +1,\,2\beta +1 \right)\] |
\[=2\alpha +1+2\beta +1\] |
\[=2\left( \alpha +\beta +1 \right)\] |
\[=2\left( -1+1 \right)=0\] |
Product of zeroes i.e., \[\left( 2\alpha +1,\,2\beta +1 \right)\] |
\[=\left( 2\alpha +1 \right)\left( 2\beta +1 \right)\] |
\[=4\alpha \beta +2\left( \alpha +\beta \right)+1\] |
\[=4\left( -2 \right)+2\left( -1 \right)+1\] |
\[=-8-2+1=-9\] |
The quadratic polynomial whose zeroes are |
\[\left( 2\alpha +1 \right),\,\left( 2\beta +1 \right)\]are |
\[={{x}^{2}}\] - (sum of zeroes) x + product of zeroes |
\[={{x}^{2}}-0\,\,\times \,\,x+\left( -9 \right)={{x}^{2}}-9\] |
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