Question

How do you solve ${x^2} + 36 = 12x$?

Answer 1

Hint: Here we will rearrange the given equation and will solve the equation as a quadratic equation and will simplify the equation by finding the roots. Finally we get the required answer.

Complete step-by-step solution:
We have the given equation:
$ \Rightarrow {x^2} + 36 = 12x$
Now on transferring terms and rearranging we get:
$ \Rightarrow {x^2} - 12x + 36 = 0$
Now since the above equation is in the quadratic format, we will find the value of $x$ as by splitting the equation, the above equation can be split up as:
$ \Rightarrow {x^2} - 6x - 6x + 36 = 0$
Now the above equation can be grouped as:
$ \Rightarrow x(x - 6) - 6(x - 6) = 0$
Since the term $(x - 6)$ is same in both the terms, we can take it out as common and write it as:
$ \Rightarrow (x - 6)(x - 6) = 0$
The above term can be written as:
$ \Rightarrow {(x - 6)^2} = 0$
Now on taking the square root on both the sides we get:
$ \Rightarrow x - 6 = 0$
On rearranging the equation, we get:

$x = 6$, which is the required answer.

Note: To find whether the value of $x$ is correct, we will substitute it in the given equation and equate it.
$ \Rightarrow {x^2} + 36 = 12x$
On substituting $x = 6$ in the left-hand side we get:
$ \Rightarrow {(6)^2} + 36$
On squaring we get:
$ \Rightarrow 36 + 36$
On simplifying we get:
$ \Rightarrow 72$
Now, on substituting $x = 6$ in the right-hand side we get:
$ \Rightarrow 12(6)$
Let us multiply we get,
$ \Rightarrow 72$
Since the left-hand side equals to the right-hand side, we can conclude that the answer is correct.
A quadratic equation is a polynomial equation with a degree $2$, quadratic equations are used mostly in statistics when there is a power.
The roots of a quadratic equation can be found using the formula $(x,y) = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2ac}}$
Where $(x,y)$ are the roots of the equation and $a,b,c$ are the coefficients of the terms in the quadratic equation
It is not necessary that both the roots of the equation will be the same and some quadratic equation might not have proper roots.