Question

If (x^2/y^2+tx+y^2/4) is a perfect square, find the value of t.​

Answer 1

Answer:

As you should be knowing (a+b)^2 = a^2 + 2ab + b^2

The Right hand side of above formula looks similar to the equation in question x^2/y^2+tx+(y^2)/4 looks similar to.

So, for comparison let us consider that

a^2 = x^2/y^2

and

b^2 = (y^2)/4

From above two, we get a= x/y & b = y/2

Let us place replace a & b into (a+b)^2 = a^2 + 2ab + b^2

We get (x/y+y/2)^2 = (x/y)^2 + 2x/y*y/2 + (y/2)^2

= x^2/y^2+x+(y^2)/4 comparing this with the question x^2/y^2+tx+(y^2)/4, we can see that

tx = x

So, t = 1