Question

Solve the given quadratic equation \[{x^2} - 36 = 0\] .

Answer 1

Hint: We will solve this question by using the quadratic formula for finding the roots which state that for any given function
\[f\left( x \right) = a{x^2} + bx + c\] , the formula for finding the roots of the function is given below:
\[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\] , where
\[a\] is the coefficient of the first term \[{x^2}\] ,
\[b\] is the coefficient of the second term \[x\] and \[c\] is constant.

Complete step-by-step solution:
Step 1: By using the quadratic formula \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\] , first of all, we will identify the value of
\[a\], \[b\] and \[c\] as given below:
\[a = 1\] , \[b = 0\] and \[c = - 36\]
Step 2: By substituting the values of
\[a = 1\] , \[b = 0\] and \[c = - 36\] in the formula \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\] , we get:
\[ \Rightarrow x = \dfrac{{ - 0 \pm \sqrt {{{\left( 0 \right)}^2} - 4\left( 1 \right)\left( { - 36} \right)} }}{{2 \times 1}}\]
By multiplying the terms inside the root of the expression \[x = \dfrac{{ - 0 \pm \sqrt {{{\left( 0 \right)}^2} - 4\left( 1 \right)\left( { - 36} \right)} }}{{2 \times 1}}\] , we get:
\[ \Rightarrow x = \dfrac{{ - 0 \pm \sqrt {0 + 144} }}{2}\]
As we know the root of \[144\] is
\[12\] , so by substituting this value in the above expression, we get:
\[ \Rightarrow x = \dfrac{{ - 0 \pm 12}}{2}\]
Step 3: Now for solving the variable, we will write the above expression in two different terms as shown below:
\[ \Rightarrow x = \dfrac{{ - 0 + 12}}{2}\] and
\[x = \dfrac{{ - 0 - 12}}{2}\]
By solving the expression \[x = \dfrac{{ - 0 + 12}}{2}\], we will calculate the first value
\[x\] as shown below:
\[ \Rightarrow x = \dfrac{{12}}{2}\]
By dividing the terms on the RHS side of the above expression, we get:
\[ \Rightarrow x = + 6\]
By solving the expression
\[x = \dfrac{{ - 0 - 12}}{2}\], we will calculate the second value \[x\] as shown below:
\[ \Rightarrow x = \dfrac{{ - 12}}{2}\]
By dividing the terms on the RHS side of the above expression, we get:
\[ \Rightarrow x = - 6\]

\[\therefore \] The values of \[x\] are \[6\] and \[ - 6\].

Note: Students can also solve this question by using the formula of \[\left( {{a^2} - {b^2}} \right) = \left( {a + b} \right)\left( {a - b} \right)\], as shown below:
We can write the equation
\[{x^2} - 36 = 0\], \[{x^2} - {6^2} = 0\].
By comparing the equation
\[{x^2} - {6^2} = 0\] with the formula \[\left( {{a^2} - {b^2}} \right) = \left( {a + b} \right)\left( {a - b} \right)\], we get:
\[ \Rightarrow \left( {{x^2} - {6^2}} \right) = \left( {x + 6} \right)\left( {x - 6} \right)\]
For finding the value of
\[x\], we will solve \[\left( {x + 6} \right) = 0\] and
\[\left( {x - 6} \right) = 0\], separately. From where we get the value of \[x = 6, - 6\].