A) \[\frac{3}{2},\frac{3}{2}\]
B) \[-\frac{3}{2},-\frac{3}{2}\]
C) \[3,4\]
D) \[-3,-4\]
Correct Answer: A
Solution :
| [a] \[f(x)=4{{x}^{2}}-12x+9\] |
| For zeroes, \[4{{x}^{2}}-12x+9=0\] |
| \[\Rightarrow \,\,\,\,\,\,{{(2x)}^{2}}-2\times 2x\times 3+{{(3)}^{2}}=0\] |
| \[\Rightarrow \,\,\,\,\,\,{{(2x-3)}^{2}}=0\] |
| \[\Rightarrow \,\,\,\,\,\,2x-3=0\] or \[2x-3=0\] |
| \[\Rightarrow \,\,\,\,\,\,x=\frac{3}{2}\] or \[x=\frac{3}{2}\] |
| \[\therefore \] Zeroes are \[\frac{3}{2},\frac{3}{2}\] |
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