Question

Find two consecutive positive odd integers have a product of 99. Find the integers

Answer 1

Answer:

Let the first odd number be

$=x$

Let the second odd number be

$x+2$

The product of the consecutive odd numbers is 99 and this will be represented below as

$x(x+2)=99$

By expanding the bracket, we will have

$\begin{gathered} x(x+2)=99 \\ x^2+2x=99 \\ x^2+2x-99=0 \end{gathered}$

By using the quadratic formula, we will have that

$\begin{gathered} x=\frac{-b\pm\sqrt{b^2-4ac}}{2a} \\ a=1,b=2,c=-99 \end{gathered}$

By substituting the values in the formula, we will have

$\begin{gathered} x=\frac{-b\pm\sqrt{b^2-4ac}}{2a} \\ x=\frac{-2\pm\sqrt{(-2)^2-4\times1\times(-99)}}{2\times1} \\ x=\frac{-2\pm\sqrt{4+396}}{2} \\ x=\frac{-2\pm\sqrt{400}}{2} \\ x=\frac{-2\pm20}{2} \\ x=\frac{-2+20}{2},x=\frac{-2-20}{2} \\ x=\frac{18}{2},x=-\frac{22}{2} \\ x=9,x=-11 \end{gathered}$

Substitute x=9 in the expression below

$\begin{gathered} x+2 \\ =9+2 \\ =11 \end{gathered}$

Hence,

The consecutive positive odd integers are

$\Rightarrow9,11$