Question

Divide 39 into two parts such that their product is 324.

Answer 1

In this problem we need to divide the given number such that the product of the two numbers is equal to$324$. We will first assume that the number $39$ is divided into $x$ , $y$ . Here the sum of the values $x$ , $y$ is equal to $39$ and the product of the values $x$ , $y$ is equal to $324$. Now we will convert them into mathematical representation and calculate the value of $y$ from any one of the equations and substitute that value in another equation and simplify it by using basic mathematical operations. Then we will get a quadratic equation, we will solve the quadratic equation by using the quadratic formula $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ and simplify the equation to get the required value. Here we will get the value of one number and the value of the second number is calculated by using any one of the equations we have.

Complete step-by-step solution:
Let us assume the two values $x$ , $y$.
The sum of the two values $x$ , $y$ is $39$.
The product of the two values $x$ , $y$ is $324$.
We can represent the both the statements mathematically as
$x+y=39$ , $xy=324$ .
Considering the equation $x+y=39$. From this equation we can write the value of $y$ as
$y=39-x$
Substituting this value in the equation $xy=324$, then we will have
$x\left( 39-x \right)=324$
Simplifying the above equation by using distribution law of multiplication, then we will get
$\begin{aligned} & 39x-{{x}^{2}}=324 \\\ & \Rightarrow {{x}^{2}}-39x+324=0 \\\ \end{aligned}$
The above equation is a quadratic equation in terms of $x$ . Comparing the above equation with $a{{x}^{2}}+bx+c=0$ , then we will get
$a=1$ , $b=-39$ , $c=324$ .
Solving the quadratic equation by using the quadratic formula $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$, then we will have
$x=\dfrac{-\left( -39 \right)\pm \sqrt{{{\left( -39 \right)}^{2}}-4\left( 1 \right)\left( 324 \right)}}{2\left( 1 \right)}$
Simplifying the above equation by using basic mathematical operations, then we will get
$\begin{aligned} & x=\dfrac{39\pm \sqrt{1521-1296}}{2} \\\ & \Rightarrow x=\dfrac{39\pm \sqrt{225}}{2} \\\ & \Rightarrow x=\dfrac{39\pm 15}{2} \\\ & \Rightarrow x=\dfrac{39+15}{2}\text{ or }\dfrac{39-15}{2} \\\ & \Rightarrow x=27\text{ or }12 \\\ \end{aligned}$
Let us assume the value of $x$ as $27$ , then the value of $y$ from the equation $y=39-x$ is given by
$\begin{aligned} & y=39-27 \\\ & \Rightarrow y=12 \\\ \end{aligned}$
Let us assume the value of $x$ as $12$ , then the value of $y$ from the equation $y=39-x$ is given by
$\begin{aligned} & y=39-12 \\\ & \Rightarrow y=27 \\\ \end{aligned}$
Hence the required two numbers which will satisfy the given conditions are $27$, $12$.

Note: In this problem we have used the quadratic formula to solve the quadratic equation. We can also use the completing squares method or the splitting the middle term method to solve the quadratic equation. In any method we will get the same solution for the quadratic equation.