one number is 5 more than half a second number, and the difference between the squares is 17. find the numbers

one number --- 2x , (avoiding fractions)

the other ----- x + 5

(2x)^2- (x+5)^2 = 17
4x^2 - x^2 - 10x -25 - 17 = 0
3x^2 - 10x - 42 = 0
x = (10 ± √604)/6
= (5 ± √151)/3

2x = ....
x+5 = ...

it works, I checked on my calculator

Let's represent the two numbers as x and y.

According to the given information, one number is 5 more than half of the second number, so we can write the first equation as:

x = (1/2)y + 5

The difference between the squares of the two numbers is 17, so we can write the second equation as:

x^2 - y^2 = 17

To solve this system of equations, we can substitute the value of x from the first equation into the second equation:

((1/2)y + 5)^2 - y^2 = 17

Expanding and simplifying this equation, we get:

(1/4)y^2 + 5y + 25 - y^2 = 17

Combine like terms:

(1/4)y^2 - y^2 + 5y + 25 = 17

Multiplying both sides of the equation by 4 to eliminate the fraction:

y^2 - 4y + 100 = 68

Rearranging the equation:

y^2 - 4y + 32 = 0

Now we can solve this quadratic equation for y by factoring, completing the square, or using the quadratic formula. However, this equation does not factor easily, so let's use the quadratic formula:

y = (-(-4) ± √((-4)^2 - 4(1)(32))) / (2(1))

Simplifying:

y = (4 ± √(16 - 128)) / 2

y = (4 ± √(-112)) / 2

Since the square root of a negative number is not a real number, there are no real solutions for y. Therefore, there are no real solutions for x either.

In conclusion, there are no real numbers that satisfy the given conditions.

To solve this problem, let's assign variables to the two numbers. Let's call the first number x and the second number y.

According to the problem, you can set up two equations:

1) "One number is 5 more than half a second number":
x = (1/2)y + 5

2) "The difference between the squares is 17":
x^2 - y^2 = 17

Now we have a system of two equations with two variables. We can solve this system using substitution or elimination.

Let's use substitution to solve:

Substitute the value of x from the first equation into the second equation:

((1/2)y + 5)^2 - y^2 = 17

Square both sides:

(1/4)y^2 + 2.5y + 5^2 - y^2 = 17

Simplify:

(1/4)y^2 - y^2 + 2.5y + 25 - 17 = 0

Combine like terms:

(-3/4)y^2 + 2.5y + 8 = 0

Multiply both sides by -4 to eliminate the fraction:

3y^2 - 10y - 32 = 0

Now we have a quadratic equation that we can solve using factoring, completing the square, or quadratic formula. Let's use factoring:

(3y + 4)(y - 8) = 0

Set each factor equal to zero and solve for y:

3y + 4 = 0 --> 3y = -4 --> y = -4/3

y - 8 = 0 --> y = 8

We have two possible values for y: y = -4/3 and y = 8.

Now substitute the values of y back into the first equation to find the corresponding values of x:

When y = -4/3:
x = (1/2)(-4/3) + 5 --> x = -2/3 + 5 --> x = 14/3

When y = 8:
x = (1/2)(8) + 5 --> x = 4 + 5 --> x = 9

So the solutions to the problem are x = 14/3 and y = -4/3, or x = 9 and y = 8.