Question

Find the numbers whose product is 30 and one is seven more than the other.

Answer 1

Hint: We will use factorization of polynomials in the form $a{{x}^{2}}+bx+c=0$ by hit and trial method. The hit and trial method is a method in which we substitute the value of x by any integer and any number that satisfies the equation $a{{x}^{2}}+bx+c=0$ will be considered as a root.
Complete step-by-step answer:
According to the question we need to find two numbers. And the relation between the two numbers is such that one number is seven more than the other. So, let us assume that one number is x and the other number is 7 more than x. This number is given by x + 7. So, the numbers are x and (x + 7). Now, according to the question these two numbers give the product as 30. Therefore, we have x (x + 7) = 30. Now, we will use multiplication to x (x + 7). Thus, we have
$\begin{align}
  & x\left( x+7 \right)=30 \\
 & \Rightarrow {{x}^{2}}+7x=30 \\
\end{align}$
Now, we will subtract 30 on both the sides. So, we have
$\begin{align}
  & {{x}^{2}}+7x-30=30-30 \\
 & \Rightarrow {{x}^{2}}+7x-30=0...(i) \\
\end{align}$
At this step we will do factorization of the equation (i) by hit and trial method. First we will substitute the value of x = 0. Therefore, we have
${{\left( 0 \right)}^{2}}+7\left( 0 \right)-30=-30$
This value of x = 0 does not satisfy the equation as $-30\ne 0$. So, now we will substitute x = 3. Therefore, we have
$\begin{align}
  & {{\left( 30 \right)}^{2}}+7\left( 3 \right)-30=9+21-30 \\
 & \Rightarrow 9+21-30=30-30 \\
 & \Rightarrow 30-30=0 \\
\end{align}$
Since, x = 3 satisfies the equation (i). Therefore, we have (x - 3) as a factor.
$\Rightarrow {{x}^{2}}+7x-30=\left( x-3 \right)\left( f\left( x \right) \right)$
To find f(x) we will divide ${{x}^{2}}+7x-30$ by x - 3. Thus, we have
$x-3\overset{x+10}{\overline{\left){\begin{align}
  & {{x}^{2}}+7x-30 \\
 & \underline{\pm {{x}^{2}}\mp 3x} \\
 & +10x-30 \\
 & \underline{\pm 10x\mp 30} \\
 & \,\,\,\,\,\,\,\,0 \\
\end{align}}\right.}}$
So, f(x) = x + 10
$\begin{align}
  & \Rightarrow {{x}^{2}}+7x-30=\left( x-3 \right)\left( x+10 \right) \\
 & \Rightarrow \left( x-3 \right)\left( x+10 \right)=0 \\
\end{align}$
Therefore, we have x - 3 = 0 or x + 10 = 0. Therefore, we get x = 3 or x = -10.
So, if the number is 3 then the other number will be 3 + 7 = 10. And if the number is - 10 then the other number will be – 10 + 7 = - 3. In both the cases the product is 30.

Note: Alternatively we could have found out the factors of the equation (i) by the method other than hit and trial method. The other method is given as the square root formula given by $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ where the equation is $a{{x}^{2}}+bx+c=0$.