Question

The product of two consecutive odd integers is $143$. How do you find the integers?

Answer 1

Hint: As we know that consecutive odd integers are those odd integers that follow each other and they differ by $2$. If $x$ is an odd integer, then $x + 2,x + 4$ and $x + 6$ are consecutive odd integers. We should represent odd integers in symbol form, and then apply this to products of consecutive odd integers. We get from the smaller number to the bigger number by adding $2$.

Complete step by step solution:
Let us take the first odd integer be $x$ and second integer be $x + 2$. In the question it is given that their product is $143$. Now by applying the same, we get the equation that:
$x(x + 2) = 143$,$ \Rightarrow {x^2} + 2x = 143$.
Here to make both the sides a perfect square we will add $1$ to both the sides and it gives us:
$ \Rightarrow {x^2} + 2x + 1 = 143 + 1$; further we get ${x^2} + 2x + 1 = 144$
We can see that both of them are the perfect squares now so, ${(x + 1)^2} = {(12)^2}$ ( if ${a^2} = {b^2}$ then we can say that $a = b$). Here $(x + 1) = 12$, solving this by transferring $1$ to right hand side we get,
$x = 12 - 1 \Rightarrow x = 11$, this is the value of the first odd integer now, we have a value of $x$ so the value of the second odd integer is $x + 2$ i.e. $11 + 2 = 13$.
Hence the required consecutive odd integers are $11$ and $13$.

Note: We have used the perfect square formula in the above solution in quadratic equation i.e.${(a + b)^2} = {a^2} + 2ab + {b^2}$, also we should remember the perfect square roots as $144$ is the square root of $12$. We have also applied the exponential formula of powers in the above solution which if the powers are the same then the base numbers are equal to each other.