Question

How do you solve ${{x}^{2}}-13x+36=0?$

Answer 1

Hint: We will use the concept of quadratic equation to solve the above question. We know that for the general quadratic equation $a{{x}^{2}}+bx+c=0$, roots of it is given by $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$. So, we will use the above formula to solve the above given quadratic equation ${{x}^{2}}-13x+36=0$.

Complete step by step anwer:
We can see from the question that equation ${{x}^{2}}-13x+36=0$ is a quadratic equation as it has degree 2.
And, for the general quadratic equation $a{{x}^{2}}+bx+c=0$, we know that root of it is given by $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$.
Now, we know from the question that we have to solve ${{x}^{2}}-13x+36=0$, which means we have to find the root of the given equation.
Thus, after comparing the given the quadratic equation ${{x}^{2}}-13x+36=0$ with general equation $a{{x}^{2}}+bx+c=0$, we will get:
a = 1, b = -13 and c = 36
Now, we will put the value of a, b, and c in $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$, to get our required roots.
$\Rightarrow x=\dfrac{-\left( -13 \right)\pm \sqrt{{{\left( -13 \right)}^{2}}-4\times 1\times 36}}{2\times 1}$
$\Rightarrow x=\dfrac{13\pm \sqrt{169-144}}{2}$
$\Rightarrow x=\dfrac{13\pm \sqrt{25}}{2}$
$\Rightarrow x=\dfrac{13\pm 5}{2}$
$\Rightarrow x=\dfrac{13+5}{2},\dfrac{13-5}{2}$
$\Rightarrow x=\dfrac{18}{2},\dfrac{8}{2}$
$\therefore x=9,4$
Hence, x = 9, 4 is the root of the given quadratic equation ${{x}^{2}}-13x+36=0$.
Thus, solution of ${{x}^{2}}-13x+36=0$ is x = 9, 4
This is our required answer.

Note:
Students are required to note that we can also solve the above question alternatively by using the middle term split method. In this method we know that the middle terms (i.e. term that has degree 1) can be divided into two term such that the sum of their coefficient is equal to the coefficient of x term and the product of them is equal to constant in term in the given polynomial.
So, for equation ${{x}^{2}}-13x+36=0$, we can split middle term as (-9x, -4x) as sum of both the term is equal to -13x and product of their coefficient is equal to $\left( -9 \right)\times \left( -4 \right)=36$ which is constant term.
$\begin{align}
  & \Rightarrow {{x}^{2}}-13x+36=0 \\
 & \Rightarrow {{x}^{2}}-9x-4x+36=0 \\
 & \Rightarrow x\left( x-9 \right)-4\left( x-9 \right)=0 \\
 & \Rightarrow \left( x-4 \right)\left( x-9 \right)=0 \\
 & \Rightarrow x=4,9 \\
\end{align}$