difference between 2 positive numbers is 3 and the sum of their squares is 117
Declare variables for each of the two numbers:
(y + 3)^2 + y^2 = 117
Evaluate and simplify:
y^2 + 3y + 3y + 9 + y^2 = 117
Combine like terms:
2y^2 + 6y + 9 = 117
Subtract 117 from each side:
2y^2 + 6y + 9 - 117 = 117 - 117
2y^2 + 6y - 108 = 0
This is a quadratic equation:
Solve the quadratic equation 2y2+6y-108 = 0
With the standard form of ax2 + bx + c, we have our a, b, and c values:
a = 2, b = 6, c = -108
Solve the quadratic equation 2y^2 + 6y - 108 = 0
The quadratic formula is denoted below:
y = -b ± sqrt(b^2 - 4ac)/2a
Step 1 - calculate negative b:
-b = -(6)
-b = -6
Step 2 - calculate the discriminant Δ:
Δ = b2 - 4ac:
Δ = 62 - 4 x 2 x -108
Δ = 36 - -864
Δ = 900 <--- Discriminant
Since Δ is greater than zero, we can expect two real and unequal roots.
Step 3 - take the square root of the discriminant Δ:
√Δ = √(900)
√Δ = 30
Step 4 - find numerator 1 which is -b + the square root of the Discriminant:
Numerator 1 = -b + √Δ
Numerator 1 = -6 + 30
Numerator 1 = 24
Step 5 - find numerator 2 which is -b - the square root of the Discriminant:
Numerator 2 = -b - √Δ
Numerator 2 = -6 - 30
Numerator 2 = -36
Step 6 - calculate your denominator which is 2a:
Denominator = 2 * a
Denominator = 2 * 2
Denominator = 4
Step 7 - you have everything you need to solve. Find solutions:
Solution 1 = Numerator 1/Denominator
Solution 1 = 24/4
Solution 1 = 6
Solution 2 = Numerator 2/Denominator
Solution 2 = -36/4
Solution 2 = -9
As a solution set, our answers would be:
(Solution 1, Solution 2) = (6, -9)
Since one of the solutions is not positive and the problem asks for 2 positive number, this problem has no solution
Declare variables for each of the two numbers:
- Let the first variable be x
- Let the second variable be y
- x - y = 3
- x^2 + y^2 = 117
- x = y + 3
- x^2 + y^2 = 117
(y + 3)^2 + y^2 = 117
Evaluate and simplify:
y^2 + 3y + 3y + 9 + y^2 = 117
Combine like terms:
2y^2 + 6y + 9 = 117
Subtract 117 from each side:
2y^2 + 6y + 9 - 117 = 117 - 117
2y^2 + 6y - 108 = 0
This is a quadratic equation:
Solve the quadratic equation 2y2+6y-108 = 0
With the standard form of ax2 + bx + c, we have our a, b, and c values:
a = 2, b = 6, c = -108
Solve the quadratic equation 2y^2 + 6y - 108 = 0
The quadratic formula is denoted below:
y = -b ± sqrt(b^2 - 4ac)/2a
Step 1 - calculate negative b:
-b = -(6)
-b = -6
Step 2 - calculate the discriminant Δ:
Δ = b2 - 4ac:
Δ = 62 - 4 x 2 x -108
Δ = 36 - -864
Δ = 900 <--- Discriminant
Since Δ is greater than zero, we can expect two real and unequal roots.
Step 3 - take the square root of the discriminant Δ:
√Δ = √(900)
√Δ = 30
Step 4 - find numerator 1 which is -b + the square root of the Discriminant:
Numerator 1 = -b + √Δ
Numerator 1 = -6 + 30
Numerator 1 = 24
Step 5 - find numerator 2 which is -b - the square root of the Discriminant:
Numerator 2 = -b - √Δ
Numerator 2 = -6 - 30
Numerator 2 = -36
Step 6 - calculate your denominator which is 2a:
Denominator = 2 * a
Denominator = 2 * 2
Denominator = 4
Step 7 - you have everything you need to solve. Find solutions:
Solution 1 = Numerator 1/Denominator
Solution 1 = 24/4
Solution 1 = 6
Solution 2 = Numerator 2/Denominator
Solution 2 = -36/4
Solution 2 = -9
As a solution set, our answers would be:
(Solution 1, Solution 2) = (6, -9)
Since one of the solutions is not positive and the problem asks for 2 positive number, this problem has no solution