Question

Solve the quadratic equation by factorisation method: ${x^2} - 15x + 54 = 0$

Answer 1

Hint:
We will factorise the equation by splitting the middle term. Split the term $ - 15x$ as $ - 9x + \left( { - 6x} \right)$ such that the product of these terms is the product of first and constant term. Then, take the terms common and represent the equation as a product of two factors.

Complete step by step solution:
We have to factorise the quadratic equation ${x^2} - 15x + 54 = 0$
We will split the middle term such that sum of the terms is the middle term and the product of the terms such as the product of the first term and constant term of the original equation.
Here, $ - 15x = - 9x + \left( { - 6x} \right)$ and $54{x^2} = \left( { - 6x} \right)\left( { - 9x} \right)$
Then, the equation cam be written as
${x^2} - 9x - 6x + 54 = 0$
Take 9 common from first two terms and $ - 6$ common from the last two terms.
$x\left( {x - 9} \right) + 6\left( {x - 9} \right) = 0$
Now, take $x - 9$ common from two terms,
$\left( {x + 6} \right)\left( {x - 9} \right) = 0$

Hence, the quadratic equation ${x^2} - 15x + 54 = 0$ is factorised as $\left( {x + 6} \right)\left( {x - 9} \right) = 0$

Note:
If we multiply the factors, we will get back the original quadratic equation. The factors divide the equation completely. If we equate each of the factors to 0, then the values of $x$ are the zeroes of the equation.