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.

Main content

Quadratic systems: both variables are squared

Sal solves the system y=0.5x and 2x^2-y^2=7. Created by Sal Khan.

Want to join the conversation?

  • leafers ultimate style avatar for user solo1118
    I don't know whether or not I should be posting this under this video, but I've been wondering about it for a while now. A LOOONG time ago when squares and square roots were explained, I thought that the square root of, say, 4 was just 2. But now that I am getting into more complicated things, it is never just 2 anymore it is + or - 2. Why is this? Are they different situations? Or is + or - just more correct?
    (15 votes)
    Default Khan Academy avatar avatar for user
    • blobby green style avatar for user Katie Berg
      + or - is just more correct. For example, yes the square root of 4 is 2. But, it also is -2 because -2 multiplied by -2 is 4 also. Depending on the teacher or the class, you may have to write it either way. The teacher I have always wants us to write both
      (18 votes)
  • mr pants teal style avatar for user Wrath Of Academy
    What kind of weird shape does 2x^2 - y^2 = 7 make?
    (8 votes)
    Default Khan Academy avatar avatar for user
  • mr pants pink style avatar for user KarlKarlJohn
    I disagree, that you can substitute back the values for x into either equation as Sal mentions at . If for example you substitute x = 2, into the 2x^2 - y^2 = 7 equation when you solve for y you get two possible values y = 1 and y = -1. This would indicate that both (2, 1) and (2, -1) are solutions, however (2, -1) is NOT a solution. By not plugging x = 2 back into the original equation y = 1/2 x, we have only insured our solutions satisfy the second equation. To see that this is the case plug (2,-1) into the first equation and you get -1 = 1/2*2, which leads to -1 = 1 which cannot be true. The reason why this isn't a problem in the second equation is that the y term is squared in that equation, which turns -1 into 1.
    (2 votes)
    Default Khan Academy avatar avatar for user
    • purple pi purple style avatar for user doctorfoxphd
      If you graph this, you will have a sideways-facing hyperbolic curve intersected by a straight line that passes through (-2, -1) (2,1), so those are the solutions. It is pretty straightforward as to what the solutions are, so what's happening with the math?
      My first suspicion was that there may be extraneous solutions if you substitute into the
      2x² - y² = 7

      (2nd degree relationships are notorious for that, and you always need to check all constraints to see they are satisfied.)

      rearranging, y² = 2x² - 7
      y = ± √(2x² - 7)
      For x = 2, y = ±√(2∙2² - 7) = ±√1
      For x = -2, y = ±√(2∙(-2)² - 7) = ±√1
      so y can potentially equal ±1 giving us the potential solutions of (2, 1) and (2, -1)as wll as (-2, 1) and (-2, -1)
      Always with these we would be cautious of extraneous solutions,. so we have to do a lot of checking. Can y =-1 when x = 2? If we use our original equation, the matter is obscured by the squares.
      So, we check with the other constraint, the linear equation and see if each of these really is a solution.

      (2, 1) Indeed, y = 1/2 x works with (2, 1) because 1 = (1/2)2 So (2,1) is a solution.
      (2, -1) If we try (2, -1), we get -1 =? (1/2)2 -1 ≠ 1 So (2, -1) is not a solution.
      (-2, 1) If we try this, we get 1 =? (1/2)(-2) +1 ≠ -1 So (-2, 1) is not a solution.
      (-2, -1) Again, -1 = (1/2)(-2) because -1 = -1 So (-2, -1) is the other solution.

      The linear equation is usually more straightforward in providing constraints and fewer extraneous answers.
      So, while you can substitute into either equation, you have to beware extraneous solutions and check your answers with all the original equations, because those are constraints on your answer.
      When in doubt, it is always helpful to sketch the graphs to figure out what is going on, and having an intuitive feel for what is reasonable helps you attack the next question with more confidence.
      (6 votes)
  • female robot grace style avatar for user Anna
    would angled parabolas(In other words not straight upwards or downwards or sideways) be of the form:

    a * sqrt(bx) +/- c * x^2 +/- dx +/- e(not the irrational number but rather a constant)?
    (2 votes)
    Default Khan Academy avatar avatar for user
    • aqualine ultimate style avatar for user Aaron Williams
      Consider a conventional Cartesian coordinate plane depicting a parabola of the form y=Ax²+Bx+C. Now imagine rotating every point on the plane θ degrees clockwise, resulting in a primed coordinate system depicting the exact same parabola, except this time it appears to be rotated when viewed from the reference frame of the xy coordinate plane. With respect to primed coordinates, our equation for the rotated parabola is y′=A(x′)²+B(x′)+C. In order to write this equation in terms of x and y coordinates, we will need to substitute equations of the form x′=f(x,y,θ) and y′=g(x,y,θ) into y′=A(x′)²+B(x′)+C.

      Let (x′,y′) be an arbitrary point on the primed Cartesian plane that makes an angle of Φ with the x′ axis. Since the x′ axis is θ degrees below the x axis, the angle that the point (x′,y′) makes with the x-axis is Φ-θ. Hence, the x and y coordinates corresponding to (x′,y′) are x=r′cos(Φ-θ) and y=r′sin(Φ-θ), where r′=(x²+y²)^0.5. We can also write x=r′cos(Φ)cos(θ)+r′sin(Φ)sin(θ) and y=r′sin(Φ)cos(θ)-r′sin(θ)cos(Φ) because of the trig identities cos(Φ-θ)=cos(Φ)cos(θ)+sin(Φ)sin(θ) and sin(Φ-θ)=sin(Φ)cos(θ)-sin(θ)cos(Φ).

      From basic trigonometry, it is clear that x′=r′cos(Φ) and y′=r′sin(Φ). This leads way to a nice simplification for the equations for x and y as follows: x=r′cos(Φ)cos(θ)+r′sin(Φ)sin(θ)=x′cos(θ)+y′sin(θ) and y=r′sin(Φ)cos(θ)-r′sin(θ)cos(Φ)=y′cos(θ)-x′sin(θ).
      Solving for x and y gives us x′=xcos(θ)-ysin(θ) and y′=xsin(θ)+ycos(θ).

      Finally, the rest of the derivation involves substituting x′ for xcos(θ)-ysin(θ) and y′ for xsin(θ)+ycos(θ) into y′=A(x′)²+B(x′)+C, which leads us to Acos²(θ)x²+Asin²(θ)y²-2Asin(θ)cos(θ)xy+(Bcos(θ)-sin(θ))x-(sin(θ)+cos(θ))y+C=0.

      In order to convince oneself that this formula is valid, it helps to try out some cases. Suppose you have the parabola y=x². Turning it 90 degrees clockwise brings us to x=y²; turning 180 degrees brings us to y=-x²;rotating 270 degrees brings us to x=-y²; and turning 360 degrees brings us back to y=x². To confirm these observations using the formula derived, let A=1, B=0, C=0 and θ=π/2,π,3π/2, and 2π to take the different degrees of rotation into account.
      (2 votes)
  • leaf green style avatar for user anirudhwadhwani
    Hello everyone!
    I wanted to ask whether or not there will be 4 solutions to these equations in the video?
    ie. (+2,+1) (+2,-1) (-2,+1) (-2,-1)
    (2 votes)
    Default Khan Academy avatar avatar for user
  • starky ultimate style avatar for user Patrick
    So, when a system is a linear equation and another is quadratic, that means there will be two solutions. Right?
    (1 vote)
    Default Khan Academy avatar avatar for user
    • duskpin sapling style avatar for user Vu
      Not always. If they never intersect (the line could be under the parabola) then there will be no solution. If they intersect at the vertex only then they will have only 1 solution and 2 solutions if they intersect twice.
      (3 votes)
  • piceratops ultimate style avatar for user Cason
    At , he says that you can solve by substation. What would I do if both equations had both x squared and y squared?
    (1 vote)
    Default Khan Academy avatar avatar for user
  • marcimus pink style avatar for user Priyasha Agarwal
    Could you use elimination for these problems?
    (1 vote)
    Default Khan Academy avatar avatar for user
  • orange juice squid orange style avatar for user Sidney B.
    This is all great and has helped me a lot but, I don't understand what happens when you have 3x^3 in the problem.
    Example: -3x^3+92y +2x-192=0, x=1/2y
    I do not understand what the solution would be to something like that.
    (1 vote)
    Default Khan Academy avatar avatar for user
  • duskpin ultimate style avatar for user Lucy
    How would you go about solving a system of quadratics in which one equation has an "xy" term?
    (1 vote)
    Default Khan Academy avatar avatar for user

Video transcript

What are the solutions to the system of equations y is equal to 1/2 x and 2x squared minus y squared is equal to 7. And they say solutions, because these two, if we view them as two curves, they could very well intersect with each other twice. So let's see what's going on here. We have y is equal to 1/2 x and 2x squared minus y squared is equal to 7. And the best way to approach these is to just try to substitute one constraint into the other constraint, or substitute one equation into the other one. It seems easier to substitute 1/2 x for y into this equation, because they've already solved for y here. Here, it's much harder to solve for x or y, so let's do that. Every place we see a y here, let's substitute it with this, that y must also be equal to 1/2 x, and then see if we can solve for x. So on this equation, we have 2x squared minus y squared. But now we're saying that y must also be equal to 1/2 x. And that is going to be equal to 7. Now let's see if we can solve for x doing a little bit of algebraic manipulation. So we get 2x squared minus. So 1/2 squared is 1/4, and then x squared. So we could say x squared over-- let me write that as 1/4. So let's say it's 1/4 x squared is equal to 7. So I have 2x squareds, and I subtract out a 1/4 x squared, so I'm going to have a 1 and 3/4 x squared. Or you could view this as 8/4 minus 1/4 is 7/4 x squared. 7/4 x squared is equal to 7. Multiply both sides times the reciprocal of 7/4, so 4/7. Multiply both sides by 4/7. And we get x squared is equal to 4. And so x could be positive or negative 2. It's the positive and negative square root of 4. So x is equal to the plus or minus square root of 4. x is equal to positive 2 or negative 2. Now, given that x is positive 2 or negative 2, let's substitute back into either of these equations to figure out what y is, what the corresponding y is for each of these. So if x is 2, y is going to be 1/2 of that. It's going to be 1. So we have the point 2 comma 1. All I did is say, look, x is 2. 1/2 times 2 is 1. If x is negative 2, then y is going to be 1/2 times negative 2, which is going to be negative 1. And both of these definitely satisfy this first constraint. And you can verify that they also satisfy this second constraint right over here. 2 times 2 squared is-- well, 2 squared is 4. 2 times that is 8. Minus 1 squared is 7. 2 times negative 2 squared is still-- this whole thing's going to be 8. Minus negative 1 squared still equals 7. So the solutions to the system of equation-- one is the coordinate 2 comma 1. x is 2. y is 1. The other is the coordinate negative 2, negative 1. x is negative 2, and y is negative 1.