Question

The product of two consecutive odd integers is equal to 30 more than the first. Find the integers.
This is solving quadratic word problems.

Answer 1

Answer:

The integers are

5 and 7

Step-by-step explanation:

Let

x ---> the first consecutive odd integer

x+2 ---> the second consecutive odd integer

we know that

The algebraic expression that represent this situation is

$x(x+2)=x+30$

solve for x

$x^2+2x=x+30\\x^2+2x-x-30=0\\x^2+x-30=0$

Solve the quadratic equation

The formula to solve a quadratic equation of the form

$ax^{2} +bx+c=0$

is equal to

$x=\frac{-b\pm\sqrt{b^{2}-4ac}} {2a}$

in this problem we have

$x^{2} +x-30=0$  

so

$a=1\\b=1\\c=-30$

substitute in the formula

$x=\frac{-1\pm\sqrt{1^{2}-4(1)(-30)}} {2(1)}$

$x=\frac{-1\pm\sqrt{121}} {2}$

$x=\frac{-1\pm11} {2}$

$x=\frac{-1+11} {2}=5$

$x=\frac{-1-11} {2}=-6$  ---> is not a odd integer

For x=5

The numbers are

$x=5\\x+2=7$

so

5 and 7