Question

The product of two alternate odd integers exceeds three times the smaller by 12. What is the larger number?
A.3
B.6
C.7
D.9

Answer 1

Hint: Here, we will first use the general forms of an odd integer and the alternate odd integer. Next, we will use the given condition to form a quadratic equation. Then, we shall factorize the quadratic equation to find its roots. Finally, we will select the appropriate root and find the alternate odd integers.

Complete step-by-step answer:
Let us take the smaller odd integer as \[2x + 1\], where \[x \in \mathbb{Z}\].
The next odd integer would be \[2x + 3\].
But we require the alternate odd integer, so we will take \[2x + 5\] as the alternate odd integer.
It is given that the product of \[2x + 1\] and \[2x + 5\] exceeds three times the smaller integer i.e., \[2x + 1\] by 12.
Thus, we have
\[(2x + 1)(2x + 5) = 3(2x + 1) + 12\]
Multiplying the terms in the LHS and RHS, we get
\[ \Rightarrow 2x \times (2x + 5) + 1 \times (2x + 5) = 3 \times 2x + 3 \times 1 + 12\]
On further simplification, we get
\[ \Rightarrow 4{x^2} + 10x + 2x + 5 = 6x + 3 + 12\]
Adding like terms on the LHS and RHS, we get
\[ \Rightarrow 4{x^2} + 12x + 5 = 6x + 15\]
Let us transpose the terms on the RHS to the LHS.
Hence, the quadratic equation becomes
\[\begin{array}{l} \Rightarrow 4{x^2} + (12x - 6x) + (5 - 15) = 0\\ \Rightarrow 4{x^2} + 6x - 10 = 0\end{array}\].
On the LHS, the coefficients are divisible by 2. So, dividing both sides of the equation by 2, we get
\[ \Rightarrow 2{x^2} + 3x - 5 = 0\] ………………………………………………….(1)
The obtained equation is a quadratic equation, so we will factorize equation (1).
We will do this by splitting the middle terms.
We need to find two terms whose product is \[2x \times 5x = 10{x^2}\] and sum/difference is \[3x\].
The appropriate terms are \[5x{\rm{ \,and\, }} - 2x\].
Using these terms to factorize equation (1), we get
\[ \Rightarrow 2{x^2} - 2x + 5x - 5 = 0\].
From the first two terms, we can take \[2x\] as common and from the last two terms we can take 5 as common. Hence,
\[ \Rightarrow 2x(x - 1) + 5(x - 1) = 0\].
Factoring out common terms, we get
\[ \Rightarrow (x - 1)(2x + 5) = 0\]
Now using zero product property, we get
\[ \Rightarrow (x - 1) = 0\] or \[(2x + 5) = 0\]
 \[ \Rightarrow x = 1\] or \[x = \dfrac{{ - 5}}{2}\]
Since we have to consider an odd integer, the appropriate value is \[x = 1\].
Now substituting \[x = 1\] in the expression for the assumed integers, we get
\[2x + 1 = 2\left( 1 \right) + 1 = 3\]
\[2x + 5 = 2\left( 1 \right) + 5 = 7\]
Hence, the alternate integers are 3 and 7.
The larger among 3 and 7 is 7.
Hence, the correct option is C.

Note: We can also take the alternate odd integers as \[x{\rm{ \,and\, }}x + 4\].
Using the given condition, we have
 \[x(x + 4) = 3x + 12\]
Multiplying the terms using distributive property, we get
 \[\begin{array}{l} \Rightarrow {x^2} + 4x = 3x + 12\\ \Rightarrow {x^2} + x = 12\end{array}\]
Rewriting the equation, we get
\[ \Rightarrow {x^2} + x - 12 = 0\]
Now using split middle term method, we get
\[\begin{array}{l} \Rightarrow {x^2} + 4x - 3x - 12 = 0\\ \Rightarrow x\left( {x + 4} \right) - 3\left( {x + 4} \right) = 0\end{array}\]
Factoring out common terms, we get
\[ \Rightarrow \left( {x + 4} \right)\left( {x - 3} \right) = 0\]
Using zero product property, we get
\[\begin{array}{l} \Rightarrow x + 4 = 0\\ \Rightarrow x = - 4\end{array}\]
Or
\[\begin{array}{l} \Rightarrow x - 3 = 0\\ \Rightarrow x = 3\end{array}\]
The roots of this equation are \[x = - 4\] and \[x = 3\], where both are integers.
But we need an odd integer which can only be \[x = 3\].
Hence, the required alternate odd integers are 3 and \[x + 4 = 3 + 4 = 7\].