Question

If \[{\left( {a + b} \right)^2} = 4ab\], then prove that the value of a is equal to the value of b (a = b).

Answer 1

Hint:- Let us use the identity of \[{\left( {a + b} \right)^2}\] to expand the LHS of the given equation ,now solve that above expression by shifting all terms to the LHS.

Complete step-by-step answer:
As we all know that if x and y are any numbers then,
\[{\left( {x + y} \right)^2} = {x^2} + {y^2} + 2xy\]
And, \[{\left( {x - y} \right)^2} = {x^2} + {y^2} - 2xy\]
So, as we know that we are given that,
\[{\left( {a + b} \right)^2} = 4ab\] (1)
So, now expand the LHS of equation 1. We get,
\[{a^2} + {b^2} + 2ab = 4ab\] (2)
Now subtracting 4ab from both sides of the equation 2. We get,
\[{a^2} + {b^2} - 2ab = 0\]
Now as I know that, \[{x^2} + {y^2} - 2xy = {\left( {x - y} \right)^2}\]. So, above equation becomes,
\[{\left( {a - b} \right)^2} = 0\] (3)
As we can see that the RHS of equation 3 is equal to zero. So, LHS must be equal to zero.
And as we know that if the square of product of two numbers is equal to zero. Then the numbers must be the same (i.e. a = b).
Hence if \[{\left( {a + b} \right)^2} = 4ab\] then a = b.
Note:- Whenever we come up with this type of problem then we should use the identity for the expansion that \[{\left( {x + y} \right)^2} = {x^2} + {y^2} + 2xy\] and \[{\left( {x - y} \right)^2} = {x^2} + {y^2} - 2xy\]. After that we will solve the equation. And we should remember that if the square of two numbers is equal then the numbers must be equal to zero, because if we subtract equal numbers then the result will be zero and the square of zero is zero itself.