Hi,
I need to solve two quadratics simultaneously to find the x and y values,
they are:
x^2 + y^2 = 0.23
(x-0.15)^2 + (y-0.175)^2 = 0.16
Thank you
Start with the second equation and expand it out:
Now substitute in for :
Now solve for y in terms of x and substitute y = f(x) back into the first equation:
Now you have a quadratic equation in terms of x which can be solved using the quadratic formula.
Forming an equation in y:
0.35y = 0.123125 - 0.3x
y =(0.123125 - 0.3x)/0.35
Substitute into x^2 + y^2 = 0.23 gives:
x^2 + ((0.123125 - 0.3x)/0.35)^2 = 0.23
Expanding brackets:
x^2 + (0.09x^2 - 0.6x + 0.015159765)/(0.35^2) = 0.23
Multiplying both sides by (0.35^2):
0.1225x^2 + 0.09x^2 - 0.6x + 0.015159765 = 0.028175
Forming quadratic equation:
0.2125x^2 - 0.6x - 0.013015235 = 0
Solving with quadratic formula:
a = 0.2125
b = -0.6
c = -0.013015235
b^2 = 0.36
4ac = -0.011062949
b^2 - 4ac = 0.371062949
sqrt(b^2 - 4ac) = 0.609149365
-b - ans = -0.009149365
-b + ans = 1.209149366
x = 2.845057332
or
x = -0.021527917
Substituting back into first equation:
when x = 2.845057332, y = NaN
when x = -0.021527917, y = 0.4790997272900794
Checking in equation 2:
(-0.021527917 - 0.15)^2 + (0.4790997272900794 - 0.175)^2 = 0.12189847124672402
The answer doesnt equal 0.16, can someone please explain what I have done wrong?