Solving Quadratic Systems
2x^2+5y^2=53
4x+y^2=17
Please show work I'm having trouble solving it.
we see that y^2 = 17-4x, so
2x^2 + 5(17-4x) = 53
2x^2 - 29x + 32 = 0
2(x-2)(x-8) = 0
...
Where did you get 29 from?
And where did you get 32 from?
4666 silly willy 420
To solve the quadratic system
2x^2 + 5y^2 = 53 .....(1)
4x + y^2 = 17 .....(2)
you can follow these steps:
Step 1: Choose one of the equations to solve for a variable in terms of the other variable. Let's solve equation (2) for y^2:
From equation (2), we have:
y^2 = 17 - 4x .....(3)
Step 2: Substitute the expression for y^2 from equation (3) into equation (1). This will give you a quadratic equation in terms of x:
Substitute y^2 = 17 - 4x into equation (1), we get:
2x^2 + 5(17 - 4x) = 53
Step 3: Simplify and rearrange the equation to the standard form: ax^2 + bx + c = 0. This will allow us to solve for x by factoring, completing the square, or using the quadratic formula:
2x^2 + 85 - 20x = 53
2x^2 - 20x + 32 = 0 .....(4)
Step 4: Solve equation (4) for x. You can do this by factoring, completing the square, or using the quadratic formula. In this case, factoring should work:
Factoring equation (4), we get:
(2x - 4)(x - 8) = 0
Setting each factor equal to zero, we have:
2x - 4 = 0 --> x = 2
x - 8 = 0 --> x = 8
Step 5: Substitute the values of x into equation (2) to find the corresponding y-values:
Substitute x = 2 into equation (2):
4(2) + y^2 = 17
8 + y^2 = 17
y^2 = 17 - 8
y^2 = 9
y = ±√9
y = ±3
Similarly, substitute x = 8 into equation (2):
4(8) + y^2 = 17
32 + y^2 = 17
y^2 = 17 - 32
y^2 = -15
Since the square of a real number cannot be negative, there are no real solutions for y.
Therefore, the solution to the quadratic system 2x^2 + 5y^2 = 53 and 4x + y^2 = 17 is (x, y) = (2, 3) and (8, no real solution).