Question

17. Write a quadratic equation to find two consecutive odd natural numbers whose
product is 63. Then find the numbers

Answer 1

Answer:

7 and 9

Step-by-step explanation:

So we want two consecutive odd numbers whose product is 63.

Let's write an equation.

Let's let n be a random integer: doesn't matter what it is. Therefore, the first integer must be 2n+1.

This is because we're letting n be whatever it wants to be. If we multiply that whatever number by 2, then it will turn even. If we add 1 to an even number, it becomes odd.

Therefore, our first odd number is (2n+1). Our second, then, must be (2n+3).

Multiply them together. They equal 63. Thus:

$(2n+1)(2n+3)=63$

Expand:

$4n^2+6n+2n+3=63$

Combine like terms:

$4n^2+8n+3=63$

Subtract 63 from both sides:

$4n^2+8n-60=0$

Divide both sides by 4:

$n^2+2n-15=0$

And now, factor:

$(n+5)(n-3)=0$

Zero Product Property:

$n+5=0\text{ or } x-3=0$

Find n:

$n=-5\text{ or } n=3$

So, we've found n.

Then the first integer is either:

$2(-5)+1 \text{ or } 2(3)+1$

Evaluate:

$-9 \text{ or } 7$

However, we want two consecutive odd natural numbers. So, ignore the -9.

Therefore, our first odd integer is 7.

And our second one would be 9.

So, our answer is 7 and 9.

And we're done!