Solving Quadratic Systems
x^2+y^2=2
x^2-6y=10
I'm having trouble solving this problem.
To solve this system of equations, we can use two different methods: the substitution method and the elimination method. Let's go through both of them step by step.
Method 1: Substitution Method
Step 1: Solve one equation for one variable.
Let's solve the second equation for x:
x^2 - 6y = 10
x^2 = 6y + 10
x = √(6y + 10)
Step 2: Substitute the value of x into the other equation.
Now, substitute the value of x into the first equation:
x^2 + y^2 = 2
(√(6y + 10))^2 + y^2 = 2
6y + 10 + y^2 = 2
Step 3: Rearrange the equation and solve for y.
Rearrange the equation by moving all terms to one side:
y^2 + 6y + 10 - 2 = 0
y^2 + 6y + 8 = 0
Step 4: Solve the quadratic equation.
To solve the quadratic equation, you can factor it or use the quadratic formula. In this case, the equation cannot be factored easily, so we'll use the quadratic formula:
y = (-b ± √(b^2 - 4ac)) / (2a)
For our equation, the coefficients are:
a = 1, b = 6, c = 8
Substituting these values into the quadratic formula:
y = (-6 ± √(6^2 - 4(1)(8))) / (2(1))
Simplifying further:
y = (-6 ± √(36 - 32)) / 2
y = (-6 ± √4) / 2
y = (-6 ± 2) / 2
y = -4/2 or y = -8/2
y = -2 or y = -4
Step 5: Substitute the values of y into the equation to solve for x.
Substitute each value of y back into the equation we obtained in Step 1:
For y = -2:
x = √(6(-2) + 10)
x = √(-12 + 10)
x = √(-2)
No real solutions exist for this case.
For y = -4:
x = √(6(-4) + 10)
x = √(-24 + 10)
x = √(-14)
No real solutions exist for this case either.
Therefore, this system of equations has no real solutions.
Method 2: Elimination Method
Step 1: Multiply one of the equations by a suitable constant to create a cancellation.
Let's multiply the first equation by 2 to make the coefficients of x^2 the same:
2(x^2 + y^2) = 2(2)
2x^2 + 2y^2 = 4
Step 2: Rearrange the equations to eliminate either x or y.
Rearrange the second equation to fit the form needed for elimination:
x^2 - 6y = 10
x^2 = 6y + 10
Step 3: Subtract the rearranged equation from the first equation.
2x^2 + 2y^2 - (6y + 10) = 4
2x^2 + 2y^2 - 6y - 10 = 4
Step 4: Simplify and solve for x.
2x^2 + 2y^2 - 6y - 10 - 4 = 0
2x^2 + 2y^2 - 6y - 14 = 0
Step 5: Determine if the equation has any real solutions.
This equation does not give us any values for x or y which satisfy the equation.
Therefore, this system of equations has no real solutions.
In conclusion, the system of equations has no real solutions, as determined by both the substitution method and the elimination method.