Solving Quadratic Systems
16x^2-9y^2=144
y=9-x^2
Please show work I'm having trouble solving it.
if you substitute y = 9-x^2, you have
16x^2 - 9(9-x^2)^2 = 144
9x^4 - 178 x^2 + 873 = 0
(x^2-9)(9x^2-97) = 0
Bot those quadratics are easy to solve, no? Each is the difference of two squares.
The correct answer is (3,0) and (3.28,-1.76)
It's correct or not
To solve the given quadratic system:
1. Start by rearranging the second equation to express y in terms of x:
y = 9 - x^2
2. Substitute this expression for y in the first equation:
16x^2 - 9(9 - x^2)^2 = 144
3. Simplify the equation by expanding and combining like terms:
16x^2 - 9(81 - 18x^2 + x^4) = 144
16x^2 - 729 + 162x^2 - 9x^4 = 144
4. Rearrange the equation by bringing all the terms to one side to form a polynomial:
9x^4 - 178x^2 + 873 = 0
5. This equation can be solved by factoring, if possible, or by using the quadratic formula. Unfortunately, this equation is a quartic equation (i.e., its highest power term is x^4), and factoring a quartic equation can be quite challenging. Therefore, to solve this equation, we'll use the quadratic formula.
6. Let u = x^2, and rewrite the equation in terms of u:
9u^2 - 178u + 873 = 0
7. Apply the quadratic formula to solve for u:
u = [-(-178) ± √((-178)^2 - 4(9)(873))] / (2 * 9)
8. Simplify the equation:
u = [178 ± √(31684 - 31332)] / 18
= [178 ± √352] / 18
9. Simplify further:
u = [178 ± 2√88] / 18
10. Reduce the fraction:
u = [89 ± √(4*22)] / 9
= [89 ± 2√22] / 9
11. Now, solve for x by taking the square root of both sides:
x = ±√((89 ± 2√22) / 9)
12. Finally, substitute the values of x into the second equation:
y = 9 - x^2
Therefore, the solution of the quadratic system is the set of ordered pairs (x, y) that satisfy both equations.