View Full Version : Non Linear System of Equations: x^2 + y^2 = 16, y = x - 3

Astark

10-30-2008, 05:18 PM

how many real solutions does this nonlinear system have:

[x]2 + [y]2 = 16

the x and y are squared

y=x-3

I've tried to use substitution, but this is not working out.

[x]2 + [x-3]2 = 16

[x]2 + [x]2 - 6x + 9 = 16

2[x]2 - 6x +9 - 16 = 0

2[x]2 - 6x - 7 = 0

if I try to factor out here, does not work?

Subhotosh Khan

10-30-2008, 10:03 PM

how many real solutions does this nonlinear system have:

[x]2 + [y]2 = 16

the x and y are squared

y=x-3

I've tried to use substitution, but this is not working out.

[x]2 + [x-3]2 = 16

[x]2 + [x]2 - 6x + 9 = 16

2[x]2 - 6x +9 - 16 = 0

2[x]2 - 6x - 7 = 0

if I try to factor out here, does not work?

What do you mean by "does not work"?

arthur ohlsten

10-30-2008, 10:48 PM

I find it helpful to sketch the problem so I can "see" the problem.

x^2+y^2=16 a circle centered at 0,0 with radius 4

y=x-3 a straight line slope 1, x intercept 3 , y intercept -3

when I sketched the equations I observed 2 interseptions

y=x-3

y^2=x^2-6x+9 substitute

x^2+[x^2-6x+9] =16

2x^2-6x-7=0

x=6+/-[36+56]^1/2 all over 4

x=6+/- 92^1/2 all over 4

x=[6+2sqrt 23]/4 answer

x=[ 6 - 2 sqrt 23]/4 answer

you can determine the y values

please check for errors

Arthur

Denis

10-31-2008, 08:58 AM

2[x]2 - 6x - 7 = 0

if I try to factor out here, does not work?

USE ^ to represent power: 2x^2 - 6x - 7 = 0

Now, do a google search on "quadratic equation": that's your "saviour" when you can't factor :idea:

arthur ohlsten

10-31-2008, 11:09 AM

factors of :

2x^2-6x-7=0

factors of 2=2,1

factors of 7=7,1

to factor the equation we want a set of factors whase difference of products = 6

none satisfy the requirement. That is why you couldnt factor.

use the binomial theorem

ax^2+bx+c=0 has the roots,[or factors ] of

x=-b+/- sqrt[b^2-4ax] all over 2a

factors are x - [3+sqrt 23] / 2

and x - [3-sqrt 23] / 2

sqrt23 is a irrational number and can't be written as a fraction, or ratio of integers .

The above is why you couldn't factor the polynomial.

Arthur

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