[SOLVED] Two numbers differ by 6. If the numbers are squared then added the sum is 146. Determine the numbers.

Realtomcruise62

New member
Joined
Feb 5, 2023
Messages
4
Problem is:

Two numbers differ by 6. If the numbers are square and then added, the sum if 146. Determine the numbers.

My working is:

Let x be the greater number and y the lesser
x - 6 = y
x^2 + y^2 = 146

using substitution:
x^2 + (x - 6)^2 = 146

Then expanding:
2x^2 - 12x + 36 = 146
2x^2 - 12x - 110 = 0
2(x^2 - 6x - 55) = 0
2(x - 11)(x + 5) = 0
Thus x is 11 and -5, but the difference between the two is 16, not 5
Where did I go wrong, thanks in advance

EDIT: I am a retard, here’s the piece I forgot about:
x = 11 or -5

x = 11
11 - 6 = y
y = 5

or

x = -5
-5 - 6 = y
y = -11

Therefore the two numbers are 11 and 5 OR -5 and -11

I assumed the roots of the quadratic were different variables for whatever reason, first time I’ve seen multiple solutions as part of a quadratic in a systems of equations problem. Mods leave this thread up for archival purposes, this seems to be nice site here and I may offer help in return at some point if I ever gain some more competence lmao, seems like a good bit of extra practice.
 
Last edited by a moderator:
Let's use algebra to solve this problem.

Let x be one of the numbers. Then the other number is x + 6 (since the two numbers differ by 6).

If we square each number and add them, we get the equation:

x^2 + (x + 6)^2 = 146

Expanding the right side of the equation, we get:

x^2 + x^2 + 12x + 36 = 146

Combining like terms, we have:

2x^2 + 12x - 110 = 0

Dividing both sides by 2, we get:

x^2 + 6x - 55 = 0

Now we can use the quadratic formula to solve for x:

x = (-6 ± sqrt(6^2 - 4(1)(-55))) / (2(1))

x = (-6 ± sqrt(256)) / 2

x = (-6 ± 16) / 2

So the possible values of x are:

x = 5 or x = -11

Since the two numbers differ by 6, the other number is either 5 + 6 = 11 or -11 + 6 = -5.

Therefore, the two numbers are either 5 and 11, or -11 and -5.
 
Let x be one of the numbers. Then the other number is x + 6 (since the two numbers differ by 6).

x^2 + (x + 6)^2 = 146

Expanding the right side of the equation, we get:

x^2 + x^2 + 12x + 36 = 146

Combining like terms, we have:

2x^2 + 12x - 110 = 0

Dividing both sides by 2, we get:

x^2 + 6x - 55 = 0

Now we can use the quadratic formula to solve for x:

x = (-6 ± sqrt(6^2 - 4(1)(-55))) / (2(1))

. . .

So the possible values of x are:

x = 5 or x = -11

Since the two numbers differ by 6, the other number is either 5 + 6 = 11 or -11 + 6 = -5.

Therefore, the two numbers are either 5 and 11, or -11 and -5.

In your let statements, indicate which variable represents the smaller number.
You expanded the left-hand side of the equation, not the right-hand side. That trinomial, with a leading coefficient of 1 and a constant of -55, lends itself to factoring instead of the cumbersome Quadratic Formula you wrote.

Since the numbers differ by 6, the larger number is stated first, and the smaller is stated second. Else, a different problem could have stated that the difference of the numbers is -6, for instance.

Therefore, the two numbers, which differ by 6, are either 11 and 5
(in this order), or -5 and -11 (in this order).

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

As an alternative, the algebra can be made relatively easier by letting the
smaller number be (x - 3) and the larger number be (x + 3).
 
Last edited:
Top