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2x^2 12x 2y^2 20y-4=0

2 Answers

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Simplifying

2x2 + 2y2 + 12x + -20y + -4 = 0

Reorder the terms:

-4 + 12x + 2x2 + -20y + 2y2 = 0

Solving

-4 + 12x + 2x2 + -20y + 2y2 = 0

Solving for variable 'x'.

Factor out the Greatest Common Factor (GCF), '2'.

2(-2 + 6x + x2 + -10y + y2) = 0

Ignore the factor 2.

Subproblem 1

Set the factor '(-2 + 6x + x2 + -10y + y2)' equal to zero and attempt to solve:

Simplifying

-2 + 6x + x2 + -10y + y2 = 0

Solving

-2 + 6x + x2 + -10y + y2 = 0

The solution to this equation could not be determined.

This subproblem is being ignored because a solution could not be determined.

The solution to this equation could not be determined.

User Charanjeet Kaur
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8.5k points
2 votes

Answer:

2x^2 + 12x + 2y^2 + 20y - 4 = 0 is x = -3 ± √(1 - 10y - y^2).

Explanation:

To solve the equation 2x^2 + 12x + 2y^2 + 20y - 4 = 0, we can follow these steps:

1. Combine like terms: Group the terms with the same variables together.

2x^2 + 12x + 2y^2 + 20y - 4 = 0

2. Set the equation equal to zero: Move the constant term (-4) to the other side of the equation.

2x^2 + 12x + 2y^2 + 20y = 4

3. Factor out any common factors: In this case, there are no common factors to factor out.

4. Use factoring or the quadratic formula to solve for x: Since we have a quadratic term (x^2) and a linear term (x), we can try factoring or using the quadratic formula. However, in this case, the quadratic does not factor nicely, so we will use the quadratic formula.

The quadratic formula is x = (-b ± √(b^2 - 4ac))/(2a), where a, b, and c are coefficients in the quadratic equation ax^2 + bx + c = 0.

For our equation 2x^2 + 12x + 2y^2 + 20y = 4, the coefficients are:

a = 2

b = 12

c = 2y^2 + 20y - 4

Substituting these values into the quadratic formula, we get:

x = (-12 ± √(12^2 - 4(2)(2y^2 + 20y - 4)))/(2(2))

5. Simplify the expression inside the square root:

x = (-12 ± √(144 - 16y^2 - 160y + 32))/(4)

6. Further simplify the expression inside the square root:

x = (-12 ± √(16 - 16y^2 - 160y))/(4)

x = (-12 ± √(16(1 - y^2 - 10y)))/(4)

x = (-12 ± √(16(1 - 10y - y^2)))/(4)

x = (-12 ± √(16(1 - 10y - y^2)))/(4)

x = (-12 ± 4√(1 - 10y - y^2))/(4)

7. Simplify the expression:

x = (-3 ± √(1 - 10y - y^2))

Therefore, the solution to the equation 2x^2 + 12x + 2y^2 + 20y - 4 = 0 is x = -3 ± √(1 - 10y - y^2).

User ThangLe
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