Question

How do you solve $ {x^2} - 49 = 0? $

Answer 1

Hint: The given equation in the form of a quadratic equation, we have 3 different methods to solve this equation. Firstly, by taking square root, next use the standard algebraic formula $ \left( {{a^2} - {b^2}} \right) = \left( {a - b} \right)\left( {a + b} \right) $ and also, we can also solve this by using the formula $ \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} $ . By using these methods, we can solve the question.

Complete step-by-step answer:
Method:1
Consider the given equation $ {x^2} - 49 = 0 $
Take 49 to the RHS then given equation and it can be written as
 $ \Rightarrow {x^2} = 49 $
Taking square root on both side
 $ \Rightarrow x = \pm \sqrt {49} $
As we know 49 is the square of 7 i.e., $ {7^2} = 49 $
 $ \Rightarrow x = \pm \,7 $
 $ \therefore \,\,x = 7 $ or $ x = - 7 $
Or
Method:2
To solve the equation $ {x^2} - 49 = 0 $ by using standard algebraic formula i.e, $ \left( {{a^2} - {b^2}} \right) = \left( {a - b} \right)\left( {a + b} \right) $
The given equation can be written as $ {x^2} - {7^2} = 0 $
Here a= $ x $ and b=7, then
 $ \Rightarrow \,{x^2} - {7^2} = 0 $
By using the standard algebraic formula it can be written as
    $ \Rightarrow \left( {x + 7} \right)\left( {x - 7} \right) = 0 $
 $ \Rightarrow \,\,\left( {x + 7} \right) = 0 $ or $ \left( {x - 7} \right) = 0 $
 $ \therefore \,\,x = - 7 $ or $ x = 7 $
Or
Method:3
Also, solve this problem by using the formula $ x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} $ . The standard quadratic equation is in the form $ a{x^2} + bx + c = 0 $
Consider the given quadratic equation $ {x^2} - 49 = 0 $ . When we compare the given equation to the standard form of the quadratic equation
Here a=1, b=0 and c=-49 then substituting the values to the formula it is written as
 $ \Rightarrow x = \,\,\dfrac{{ - 0 \pm \sqrt {{0^2} - 4\left( 1 \right)\left( { - 49} \right)} }}{{2.1}} $
On simplification we have
 $ \Rightarrow x = \,\,\dfrac{{ \pm \sqrt {196} }}{2} $
As we know 149 is the square of 14 i.e., $ {14^2} = 149 $
 $ \Rightarrow \,x = \dfrac{{ \pm \sqrt {{{14}^2}} }}{2} $
 $ \Rightarrow x = \dfrac{{ \pm 14}}{2} $
 $ \Rightarrow x = \pm 7 $
 $ \therefore \,\,x = 7 $ or $ x = - 7 $
By the three methods also we obtain the same solution.
Hence, after solving the given quadratic equation $ {x^2} - 49 = 0 $ we get the roots of x=7 or x=-7.
So, the correct answer is “ $ x = 7 $ OR $ x = - 7 $ ”.

Note: The given equation is a quadratic function, the general quadratic equation is represented as $ a{x^2} + bx + c = 0 $ where a, b and c are known values and x is unknown value where $ a \ne 0 $ . We can solve this quadratic equation by using the formula $ \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} $ and we can also use algebraic formulas.