Question

How do you solve \[3{{x}^{2}}+45=24x\] by factoring?

Answer 1

Hint: These are general steps to factoring simple quadratic equations:
If we have a quadratic equation of the form $a{{x}^{2}}+bx+c$,
Finding two numbers that multiply to give ac and add to give b, usually does the trick.
Here, a=3 b=-24 and c =45, but this equation is further simplified before factoring to get a=1 because its coefficients have common factors.

Complete step by step solution:
This type of question is based on the concept of factoring a quadratic equation. Factoring means finding expressions that can be multiplied together to give the expression on one side of the equation. If a quadratic equation can be factored, it is written as a product of linear terms.
Since subtracting adding, multiplying, or dividing terms on both sides of an equality does not change it:
Dividing both sides of \[3{{x}^{2}}+45=24x\] by 3 we get,
$\Rightarrow \dfrac{3{{x}^{2}}+45}{3}=\dfrac{24x}{3}$
$\Rightarrow \dfrac{3{{x}^{2}}}{3}+\dfrac{45}{3}=\dfrac{24x}{3}\Rightarrow {{x}^{2}}+15=8x$
Subtracting both sides by 6x we get:
This is done to get the equation on one side so it can be factored.
$\Rightarrow {{x}^{2}}-8x+15=0$
Now to help us guess the factors, we first list down all the numbers such that their product is 15:
$\begin{align}
  & 1\times 15 \\
 & 5\times 3 \\
 & -5\times -3 \\
\end{align}$
Now, from the above 5 and 3 will add up to 8
Thus, we have:
$\Rightarrow {{x}^{2}}-5x-3x+15=0$
Now, taking x common from the first two terms:
\[\Rightarrow x\left( x-5 \right)-5x+15\]
Now, taking -5 common from the last two terms
\[\Rightarrow x\underline{(x-5)}-3\underline{(x-5)}=0\]
Now, we take (x-5) common, and we get,
$\Rightarrow (x-5)(x-3)=0$

Thus x=3, 5 is the solution to the equation \[3{{x}^{2}}+45=24x\].

Note: It is always a good idea to take out any common factors of all the terms in an equation before trying to factorize it. This is because it reduces the value of the coefficient and for quadratic equations, that means less factors to list for ‘ac’.
Alternatively, after reaching ${{x}^{2}}-6x+9=0$ equation, one can always compare it to the formulae we know such as ${{(x-a)}^{2}}={{x}^{2}}-2ax+{{a}^{2}};{{(x+a)}^{2}}={{x}^{2}}+2ax+{{a}^{2}};(x-a)(x+a)={{x}^{2}}-{{a}^{2}}$
These usually end up giving the answer we want much faster, so it is particularly important to do just a quick check and use the above method when none of the formulae hold.