A) \[\frac{4}{3}\]
B) \[-\frac{4}{3}\]
C) \[\frac{2}{3}\]
D) \[\frac{-2}{3}\]
Correct Answer: A
Solution :
| Given that, one of the zeroes of the quadratic polynomial say \[p\left( x \right)=\left( k-1 \right){{x}^{2}}+kx+1\] is |
| \[-3\], then \[p\left( -3 \right)=0\] |
| \[\Rightarrow \,\left( k-1 \right){{\left( -3 \right)}^{2}}+k)-3+1=0\] |
| \[\Rightarrow \,9\left( k-1 \right)-3k+1=0\] |
| \[\Rightarrow \,9k-9-3k+1=0\] |
| \[\Rightarrow \,6k-8=0\] |
| \[\therefore k=\frac{4}{3}\] |
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