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

Printable View

- Feb 23rd 2010, 11:07 AMsweetiepieSimultaneous quadratic equations
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 - Feb 23rd 2010, 11:15 AMicemanfan
Start with the second equation and expand it out:

$\displaystyle (x - 0.15)^2 + (y - 0.175)^2 = 0.16$

$\displaystyle x^2 - 0.3x + 0.15^2 + y - 0.35y + 0.175^2 = 0.16$

$\displaystyle x^2 + y^2 - 0.3x - 0.35y = 0.16 - 0.15^2 - 0.175^2$

Now substitute in for $\displaystyle x^2 + y^2$:

$\displaystyle 0.23 - 0.3x - 0.35y = 0.106875$

$\displaystyle 0.3x + 0.35y = 0.123125$

Now solve for y in terms of x and substitute y = f(x) back into the first equation:

$\displaystyle x^2 + (f(x))^2 = 0.23$

Now you have a quadratic equation in terms of x which can be solved using the quadratic formula. - Feb 23rd 2010, 11:55 AMsweetiepie
Thanks for helping me but when I substitute in the results I get from the quadratic formula into the two equations I do not get the correct result,

Can someone show me how to get to the quadratic formula so I can check my answers please? - Feb 23rd 2010, 04:37 PMHallsofIvy
That should work. How about if YOU show your work and why you think the answer you got was wrong?

- Feb 24th 2010, 05:15 AMsweetiepie
**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?*