Question

How do you solve ${x^2} = 49?$

Answer 1

Hint: Here we will find the values for “x”. we can use the identity of difference between the two squares and then can find the required resultant value for “x”.

Complete step-by-step solution:
Take the given expression: ${x^2} = 49$
Take all the terms on one side. Move term from the right hand side to the left hand side. When you move any term from one side to another the sign of the terms also changes. Positive terms become negative and vice-versa.
$\Rightarrow {x^2} - 49 = 0$
Observe the above expression, it is difference between the two squares and hence it can be re-written as:
$\Rightarrow {(x)^2} - {(7)^2} = 0$
The above equation satisfies the difference of two squares identity and therefore apply ${a^2} - {b^2} = (a - b)(a + b)$in the above expression.
$\Rightarrow (x - 7)(x + 7) = 0$
Therefore, we get two values,
Case (I)
$\Rightarrow x - 7 = 0$
Move the constant term on the opposite side, when you move any term from one side to another then the sign of the term also changes. Negative term becomes positive term.
$\Rightarrow x = 7$ …. (A)
Similarly, Case (II)
$\Rightarrow x + 7 = 0$
Move the constant term on the opposite side, when you move any term from one side to another then the sign of the term also changes. Positive term becomes negative term.
$\Rightarrow x = - 7$ …. (B)

Hence, the required solution is $x = (7,- 7)$

Note: The above solution can be done using another method.
Take the given expression: ${x^2} = 49$
Take the square root on both the sides of the equation.
$\sqrt {{x^2}} = \sqrt {{{( \pm 7)}^2}} $
Square and square root cancel each other.
$ \Rightarrow x = \pm 7$
Also, remember the square of positive and the negative term gives a result always as positive and square root of the positive term can give negative or positive terms.