SOLUTION: find 2 consecutive integers such that three times the square of the first is equal to seven more than five times the second. what is the answer?

Question 225261: find 2 consecutive integers such that three times the square of the first is equal to seven more than five times the second.
what is the answer?

Answer by drj(1380) (Show Source): You can put this solution on YOUR website!
Find 2 consecutive integers such that three times the square of the first is equal to seven more than five times the second.
What is the answer?

Step 1. Let be the first integer.

Step 2. Let be the next consecutive integer

Step 3. Let be three times the square of the first.

Step 4. Let be seven more than five times the second.

Step 5. Then since three times the square of the first is equal to seven more than five times the second.

Step 6. Solving the equation in Step 5 yields the following steps.

Subtract 5n+12 from both sides of the equation

Step 7. To solve use the quadratic equation given as

where a=3, b=-5, and c=-12

Solved by pluggable solver: SOLVE quadratic equation with variable

Quadratic equation (in our case ) has the following solutons:

For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=16 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: 3, -1. Here's your graph:

With n=3 then n+1=4. Check if true so or which is a true statement.

With n=-1 then n+1=0 Check if true so or which is not a true statement.

Step 8. ANSWER: The numbers are 3 and 4

I hope the above steps were helpful.

For FREE Step-By-Step videos in Introduction to Algebra, please visit http://www.FreedomUniversity.TV/courses/IntroAlgebra and for Trigonometry visit http://www.FreedomUniversity.TV/courses/Trigonometry.

And good luck in your studies!

Respectfully,
Dr J
http://www.FreedomUniversity.TV

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