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Elden [556K]
3 years ago
8

17. Write a quadratic equation to find two consecutive odd natural numbers whose

Mathematics
1 answer:
Molodets [167]3 years ago
8 0

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 <em>must</em> 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 <em>will</em> 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 <em>natural </em>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!

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