Question

If an algebraic expression ${{\left( a+b \right)}^{2}}=4ab$ is given, then prove that $a = b$.

Answer 1

Hint: We start solving the problem by expanding ${{\left( a+b \right)}^{2}}$. After expanding ${{\left( a+b \right)}^{2}}$, we get to see a change in the value of the coefficient of ‘ab’ term. After changes in the value of the coefficient of ‘ab’ term, we can observe that the equation will convert to ${{\left( a-b \right)}^{2}}$. Using this we prove the required result.

Complete step-by-step solution:
Given that we have ${{\left( a+b \right)}^{2}}=4ab$. We need to prove $a = b$.
We have got ${{\left( a+b \right)}^{2}}=4ab$ ------(1).
We know that ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$, and let us substitute this in equation (1).
We have got ${{a}^{2}}+{{b}^{2}}+2ab=4ab$.
We have got ${{a}^{2}}+{{b}^{2}}=4ab-2ab$.
We have got ${{a}^{2}}+{{b}^{2}}=2ab$.
We have got ${{a}^{2}}+{{b}^{2}}-2ab=0$ ---------(2).
We know that ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$, and let us substitute this in equation (2).
We have got ${{\left( a-b \right)}^{2}}=0$.
We have got $a-b=\sqrt{0}$ ---(3).
We know that the square root of zero is zero. Let us use this in equation (3).
We have got $a – b = 0$.
We have got $a = b$.
$\therefore$ We have proved $a = b$.

Note: We can alternatively solve the problem as follows:
We have got ${{\left( a+b \right)}^{2}}=4ab$
We have got $a+b=\sqrt{4ab}$.
We know that $\sqrt{abc}=\sqrt{a}.\sqrt{b}.\sqrt{c}$.
We have got $a+b=\sqrt{4}.\sqrt{a}.\sqrt{b}$.
We have got $a+b=2.\sqrt{a}.\sqrt{b}$ --------(4).
We know that ${{\left( \sqrt{x} \right)}^{2}}=x$ and we use this in equation (4).
We have got ${{\left( \sqrt{a} \right)}^{2}}+{{\left( \sqrt{b} \right)}^{2}}=2\sqrt{a}\sqrt{b}$.
We have got ${{\left( \sqrt{a} \right)}^{2}}+{{\left( \sqrt{b} \right)}^{2}}-2\sqrt{a}\sqrt{b}=0$ ----------(5).
We know that ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$, and let us substitute this in equation (5).
We have got ${{\left( \sqrt{a}-\sqrt{b} \right)}^{2}}=0$.
We have got $\sqrt{a}-\sqrt{b}=0$.
We have got $\sqrt{a}=\sqrt{b}$.
Now we square on both sides.
We have got ${{\left( \sqrt{a} \right)}^{2}}={{\left( \sqrt{b} \right)}^{2}}$------(6).
We know that ${{\left( \sqrt{x} \right)}^{2}}=x$ and we use this in equation (6).
We have got $a = b$.

$\therefore$ We have proved $a=b$.