Question

Solve $ {x^2} = 0 $

Answer 1

Hint: The given problem comprises a quadratic equation in one variable. There are various methods to solve a quadratic equation like factorisation method, completing the square method, splitting the middle term and using quadratic formula. The given equation can be solved easily by using simple algebraic rules like transposition and algebraic identities.

Complete step-by-step answer:
The given quadratic equation in one variable has no linear and constant term.
So, $ {x^2} = 0 $
Now, we transpose the zero to the left side of the equation.
 $ = $ $ {x^2} - 0 = 0 $
Now, we write $ 0 $ as $ {0^2} $ , to make the left side of equation resemble $ ({a^2} - {b^2}) $
 $ = $ $ \left( {{x^2}} \right) - {\left( 0 \right)^2} = 0 $ $ $
Now, using the algebraic identity $ \left( {{a^2} - {b^2}} \right) = \left( {a - b} \right)\left( {a + b} \right) $ , we get,
 $ = $ $ \left( {x + 0} \right)\left( {x - 0} \right) = 0 $ $ $
Now, either \[\left( {x + 0} \right) = 0\] or $ \left( {x - 0} \right) = 0 $
So, either $ x = 0 $ or $ x = 0 $
Hence, $ x = 0 $
So, the correct answer is “ $ x = 0 $ ”.

Note: The given quadratic equation $ {x^2} = 0 $ can also be solved with pure logic as the equation given is simple and straightforward. We know that square of any real number is always greater than or equal to $ 0 $ . Also, the square of any real number except $ 0 $ is always greater than $ 0 $ . Hence, $ x = 0 $ as $ 0 $ is the only real number whose square is equal to $ 0 $ .
The problem can also be solved using the quadratic formula $ x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} $ where a is coefficient of $ {x^2} $ b is the coefficient of $ x $ and c is the constant term in the quadratic equation. This method is also as efficient as the former one and directs us towards the correct answer.