A) \[{{x}^{2}}+2x+24\]
B) \[{{x}^{2}}-2x-24\]
C) \[{{x}^{2}}-2x+24\]
D) None of these
Correct Answer: B
Solution :
Let \[p\left( x \right)={{x}^{2}}-5\] |
For finding the zeroes of p(x), put p(x) = 0. |
\[\therefore \,\,\,{{x}^{2}}-5=0\Rightarrow x=\pm 5\] |
Let \[\alpha =5\] and \[\beta =-5\] |
Now, \[1+\alpha =1+5=6\] and \[1+\beta =1-5=-4\] |
Thus, 6 and - 4 are the zeroes of new quadratic polynomial. Therefore the new quadratic polynomial i |
\[{{x}^{2}}\] - (sum of zeroes) x + product of zeroes |
\[\therefore\] Required polynomial |
\[={{x}^{2}}-\left[ 6+\left( -4 \right)x+6\times \left( -4 \right) \right]\] |
\[={{x}^{2}}-2x-24\] |
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