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Solve the equation ${x^2} - 12x + 27 = 0$

Answer
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Hint: Here we asked to solve the given equation for the value of $x$ .We can see that the given equation is a quadratic equation, any quadratic equation can be solved for their unknown variable by using the quadratic root formula. Since the given equation is a quadratic equation, we will get two roots for this equation because of the degree of the equation of two. The formula to solve this equation is given in the following section.

Formula:
Let the equation be in the form $a{x^2} + bx + c = 0$ then it is said to be a quadratic equation, its roots can be found by the formula: $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$

Complete step by step solution:
It is given that ${x^2} - 12x + 27 = 0$ we aim to solve this equation, that is we have to find its roots.
We know that the number of roots of an equation is equal to its degree. Here the degree of the given equation is two thus this equation will have two roots.
The roots of a quadratic equation can be found by using the formula \[\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\] where \[a\] - coefficient of the term \[{x^2}\], \[b\] - coefficient of the term \[x\], and \[c\] - constant term.
First, let us collect the required terms for the formula from the given quadratic equation to solve it.
The given equation is in the general quadratic equation form ${x^2} - 12x + 27 = 0$, from that we get \[a = 1\], \[b = 12\], and \[c = 27\].
Now let’s substitute these values in the formula to find the roots.
\[x = \dfrac{{ - \left( {12} \right) \pm \sqrt {{{\left( {12} \right)}^2} - 4\left( 1 \right)\left( {27} \right)} }}{{2\left( 1 \right)}}\]
On simplifying this we get
\[x = \dfrac{{12 \pm \sqrt {144 - 108} }}{2}\]
On further simplification we get
\[x = \dfrac{{12 \pm \sqrt {36} }}{2}\]
\[ = \dfrac{{12 \pm 6}}{2}\]
\[ \Rightarrow x = = \dfrac{{12 + 6}}{2}\] and \[ \Rightarrow x = = \dfrac{{12 - 6}}{2}\]
On solving the above, we get
\[x = 9\] and \[x = 3\]
Thus, we got the roots of the given quadratic equation that is \[x = 9\] and \[x = 3\]

Note: We can also solve the given equation by following method:
Consider the given equation ${x^2} - 12x + 27 = 0$ now let us split the middle term like
${x^2} - 9x - 3x + 27 = 0$
Now let us group the terms having common factors.
$x\left( {x - 9} \right) - 3\left( {x - 9} \right) = 0$
$\left( {x - 9} \right)\left( {x - 3} \right) = 0$
This implies $x - 9 = 0$ and $x - 3 = 0$
From the above we get $x = 9$ and $x = 3$
Hence, we got the answer.