Question

Show that if $2\left( {{a}^{2}}+{{b}^{2}} \right)={{\left( a+b \right)}^{2}}$ , then a = b.

Answer 1

Hint: To solve this question, firstly, we will expand the right hand side equation using algebraic identity ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ and solve the left hand side by solving the brackets, after that, we will move equation on right to left side and solve the equations. After that we will get the identity of ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$. And by solving the equation ${{\left( a-b \right)}^{2}}=0$, we get a – b = 0 or a = b.

Complete step-by-step solution:
Now, in question we are given that
$2\left( {{a}^{2}}+{{b}^{2}} \right)={{\left( a+b \right)}^{2}}$
We know that, ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$
So, on expanding the left hand side, we get
$2\left( {{a}^{2}}+{{b}^{2}} \right)={{a}^{2}}+{{b}^{2}}+2ab$
On solving brackets on right hand side, we get
$2{{a}^{2}}+2{{b}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$
Now, on taking all the expression on right hand side, we get
$2{{a}^{2}}+2{{b}^{2}}-\left( {{a}^{2}}+{{b}^{2}}+2ab \right)=0$
On simplifying, we get
${{a}^{2}}+{{b}^{2}}-2ab=0$
We know that, ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$
So, we can re – write above equation as
${{\left( a-b \right)}^{2}}=0$
Taking square root on both sides, we get
$\pm \left( a-b \right)=0$
So, we have two case
$a – b = 0$ or $– a + b = 0$
on simplifying, we get
$a = b$
Hence, if $2\left( {{a}^{2}}+{{b}^{2}} \right)={{\left( a+b \right)}^{2}}$ , then a = b.

Note: Always remember that when we take the square root of an expression we get two final equations for a positive and negative sign, so consider both cases. We can also solve the equation by direct substituting a = b, and on further solving we will end up with values on both sides. Try not to do any calculation and learn all algebraic identities.