If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

## Algebra (all content)

### Course: Algebra (all content)>Unit 9

Lesson 11: Systems of quadratic equations

# Quadratic systems: a line and a circle

Sal solves the system y=x+1 and x^2+y^2=25. Created by Sal Khan.

## Want to join the conversation?

• At , Sal draws a circle, but from an equation like the one in the video, is there any way to figure out exactly how large the circle is?
• You can know it by substituting y by 0, so x will have two solutions on the x-axis, one is positive and the other is negative. The same would be if you substituted x by 0, y will have two solutions on the y-axis one is positive and the other is negative.
• How does he get the 2x?
• (x+1)² = (x+1)(x+1)
Use FOIL to get (x)(x) + 1x + 1x + (1)(1) = x² + 2x + 1
• At of the video, how does Sal get that x= –4 or x= 3 from the "(x+4)(x-3)=0?"
• When we get down to the factored formula (x+4)(x-3)=0, then we use the zero property principle that the others have described to finish the solution. The next step is to set each factor equal to zero by itself and solve.
(x+4) = 0
x + 4 -4 = 0 -4
`x = -4 First solution for x`
(x-3) = 0
x-3 +3 = +3
`x = 3 Second solution for x`
• I can easily do those in my head, and so can many others. Why do we have to do all this work when It could be so much easier?
• With simple examples it may seem like we are hitting thumbtacks with a 30 pound sledgehammer rather than just pushing them in with our thumb. The point is to teach you, and get you used to, a method of solving equations, a method that works when you can't do the problems in your head because they are too complicated. You may be a fortunate person in that these processes are easy for you, but there are many others here for which this is hard - so they need this.
• Could we also use elimination like this?
``Turn y = x+1 into y^2 = x^2 + 2x + 1, Turn x^2 + y^2 = 25 into y^2 = 25 - x^2,Subtract to get (x^2 + 2x + 1) - (25-x^2) = 0,Simplify to the quadratic 2x^2 +2x - 24 = 0,``

and continuing as shown in the video?
• Yes, that works... you have the same quadratic to solve as Sal.
• why can one not simply sqrt x^2+(x+1)^2=25 and get x+x+1=5? I mean isn't the sqrt of x^2 =x and the sqrt of (x+1)^2 = x+1 and the sqrt of 25 = 5?
• √(a+b+c+...) is not equal to √(a)+√(b)+√(c)+...

So, √(x^2 + (x+1)^2) does not equal √(x^2) + √((x+1)^2)
• Hrm. I am a little disappointed to find out this only covers substitution of simpler algebraeic models. My question is: How do you solve the very most general models? Something like 3 = A(e^2B) + C, 7 = A(e^4B) + C and 9 = A(e^6B) + C using some numerical method. I don't think Khan Academy actually covers this. basically regression of nonlinear models using advanced numerical methods
• I can show you the methods for solving problems like that, if you need it. But they do get rather complicated.
• Could you please give the formula to turn "x^2 + y^2 = 25" into a circle with it's center at 0 and radius of 5? I know it's not directly related to the solution(s), but I would still like to know it. Thanks!
• You don't 'turn' that equation into a circle with radius 5 and center (0,0)... it already is that circle.

The general form for the equation of a circle is:
(x-h)^2 + (y-k)^2 = r^2 is the equation of a circle with center at (h,k) and radius r.
So, (x-4)^2 + (y+2)^2 = 49 has h=4, k=-2 and r=7, so it is the equation of a circle with center at (4, -2) and radius 7

For x^2 + y^2 = 25, think of it as (x-0)^2 + (y-0)^2 = 5^2, so it is a circle with center at (0, 0) and radius 5.