Question

Solve: ${x^2} - 6x + 9 = 0$

Answer 1

Hint:Given equation is of degree $2$. Equations of degree $2$ are known as quadratic equations. Quadratic equations can be factored by the help of splitting the middle term method. In this method, the middle term is split into two terms in such a way that the equation remains unchanged and becomes easy to factorise.

Complete step by step answer:
 For factorising the given quadratic equation ${x^2} - 6x + 9 = 0$ , we use the splitting the middle term method. Now, we have to factorise the provided quadratic equation. We can use the splitting the middle term method in which the middle term is split into two terms such that the sum of the terms gives us the original middle term and the product of the terms gives us the product of the constant term and coefficient of ${x^2}$.

So, we get,
$ \Rightarrow {x^2} - \left( {3 + 3} \right)x + 9 = 0$
We split the middle term $ - 6x$ into two terms $ - 3x$ and $ - 3x$ since the product of these terms, $9{x^2}$ is equal to the product of the constant term and coefficient of ${x^2}$ and sum of these terms gives us the original middle term, $ - 6x$.
$ \Rightarrow {x^2} - 3x - 3x + 9 = 0$
Taking x common from the first two terms and $ - 3$ common from the last two terms. We get,
$ \Rightarrow x\left( {x - 3} \right) - 3\left( {x - 3} \right) = 0$
$ \Rightarrow \left( {x - 3} \right)\left( {x - 3} \right) = 0$
$ \Rightarrow {\left( {x - 3} \right)^2} = 0$
So, the factored form of the equation ${x^2} - 6x + 9 = 0$ is ${\left( {x - 3} \right)^2} = 0$.
Now, since the square of $\left( {x - 3} \right)$ is equal to zero. So, the term $\left( {x - 3} \right)$ itself should be equal to zero.
So, the solution of the equation ${x^2} - 6x + 9 = 0$ is the same as the solution of the linear equation $\left( {x - 3} \right) = 0$.
Now, shifting the constant term to the right, we get,
$ \therefore x = 3$

Therefore, the solution of the quadratic equation ${x^2} - 6x + 9 = 0$ is $x = 3$.

Note:Splitting of middle term can be a tedious process at times when the product of the constant term and coefficient of ${x^2}$ is a large number with a large number of divisors. Special care should be taken in such cases. Similar to quadratic polynomials, quadratic solutions can also be solved using the factorisation method. Besides factorisation, there are various methods to solve quadratic equations such as completing the square method and using the Quadratic formula.