Question: The value of t for which $ m^{2}-\frac{3m}{2}+t $ will be a perfect square, is
Options:
A) $ \frac{9}{4} $
B) $ \frac{9}{16} $
C) $ \frac{3}{2} $
D) $ \frac{3}{4} $
Show Answer
Answer:
Correct Answer: B
Solution:
- $ m^{2}-\frac{3m}{2}+t={{( m-\frac{3}{4} )}^{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.