Question

Find three consecutive positive integers such that the sum of the square of the first and the product of the other two is 154.

Answer 1

Hint:
In this question, we need to find three consecutive integers if we are given a statement about their relationship. For this, we will first suppose three consecutive integers as (x-1), x, (x+1). Then we will apply the given statement to form an equation. Solving the equation will give us a value of x. Using that, we will then find the required three integers.

Complete step by step answer:
Here we need to find three consecutive positive integers. Let us suppose that, one of them is x. Since the integers are consecutive, the other two integers can be (x-1) and (x+1). Therefore, our three consecutive positive integers are (x-1), x, (x+1).
Now, we are given that the sum of the square of the first term and the product of the other two is 144. So let us first find the square of the first term. We get $ {{\left( x-1 \right)}^{2}} $ .
Using $ {{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab $ we get, $ {{x}^{2}}+1-2x $ .
Now let us find the product of the second term and the third term we get, $ x\left( x+1 \right)={{x}^{2}}+x $ .
According to question, our statement becomes that sum of $ \left( {{x}^{2}}+1-2x \right)\text{ and }\left( {{x}^{2}}+x \right) $ is equal to 154.
In mathematical form we can write it as $ {{x}^{2}}+1-2x+{{x}^{2}}+x=154 $ .
Solving and rearranging we get $ 2{{x}^{2}}-x+1-154=0\Rightarrow 2{{x}^{2}}-x-153=0 $ .
Let us apply the splitting the middle term method, we know -18+17 = -1 and (-18)(17) = -153. So we get, $ 2{{x}^{2}}-18x+17x-153=0 $ .
Taking 2x common from the first two terms and 17 common from the third and fourth term we get, $ 2x\left( x-9 \right)+17\left( x-9 \right)=0\Rightarrow \left( 2x+17 \right)\left( x-9 \right) $ .
We can say $ \begin{align}
  & 2x+17=0\text{ and }x-9=0 \\
 & \Rightarrow 2x+17=0\Rightarrow 2x=-17 \\
\end{align} $ .
Dividing by 2 on both sides we get, $ x=\dfrac{-17}{2} $ .
Also $ x-9=0\Rightarrow x=9 $ .
Since we are given positive integers so we cannot pick $ x=\dfrac{-17}{2} $ . Hence the value of x = 9.
Other two numbers are $ x-1=9-1=8\text{ and }x+1=9+1=10 $ .
So 8, 9, 10 are required to have three consecutive positive integers.

Note:
 Students can also take three consecutive integers as x, (x+1), and (x+2). Make sure to ignore the terms $ x=\dfrac{-17}{2} $ as it is neither an integer nor a positive number. Take care of the signs while applying a split middle term method.