| The value of t for which \[{{m}^{2}}-\frac{3m}{2}+t\]will be a perfect square, is |
A) \[\frac{9}{4}\]
B) \[\frac{9}{16}\]
C) \[\frac{3}{2}\]
D) \[\frac{3}{4}\]
Correct Answer: B
Solution :
| \[{{m}^{2}}-\frac{3m}{2}+t={{\left( m-\frac{3}{4} \right)}^{2}}+t-\frac{9}{16}\] |
| So, if \[t=\frac{9}{16},\]it will become a perfect square. |
| Alternate Method |
| Given equation \[={{m}^{2}}-\frac{3m}{2}+t\] |
| Now, taking the standard perfect square |
| \[{{(x-y)}^{2}}={{x}^{2}}+{{y}^{2}}-2xy\] |
| On comparing, we get |
| \[{{m}^{2}}={{x}^{2}},\]\[{{y}^{2}}=t\] and \[\frac{3}{2}m=2m\sqrt{t}\] |
| \[\Rightarrow \] \[\sqrt{t}=\frac{3}{4}\]\[\Rightarrow \]\[t=\frac{9}{16}\] |
| Hence, value of t should be \[\frac{9}{16}\] to make the given equation a perfect square. |
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