Page 1 of 2 12 LastLast
Results 1 to 15 of 16
Like Tree2Thanks

Thread: Linear Quadratic Systems Help

  1. #1
    Newbie
    Joined
    Jul 2018
    From
    Canada
    Posts
    8

    Linear Quadratic Systems Help

    Determine the restrictions on the y-intercept so that y = 3x^2 + 6x - 1 intersects with a line with slope 2 in more than one place.
    I equate the problem like how it was shown in my lesson
    and made an equation for slope 2
    y = 3x^2 +6x -1 (Quadratic Equation)
    y = 2x + b (Linear Equation)
    3x^2 + 6x -1 = 2x + b
    3x^2 + 6x - 2x -1 -b
    3x^2 + 4x -1 - b
    And I equate the problem and I get
    3x^2 + 4x - 1 - b
    I'm not sure if i'm able to use this equation in the quadratic formula since there is a,b,c and d in this equation.



    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Jan 2009
    Posts
    396
    Thanks
    67

    Re: Linear Quadratic Systems Help

    Since 1 and b are constants with respect to x, you can group them in brackets, like so
    3x^2 + 4x + (-1-b) = 0

    Can you finish?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Jul 2018
    From
    Canada
    Posts
    8

    Re: Linear Quadratic Systems Help

    Quote Originally Posted by MacstersUndead View Post
    Since 1 and b are constants with respect to x, you can group them in brackets, like so
    3x^2 + 4x + (-1-b) = 0

    Can you finish?
    Would that make it 1b?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member
    Joined
    Jan 2009
    Posts
    396
    Thanks
    67

    Re: Linear Quadratic Systems Help

    No.
    Consider the polynomial ax^2 + bx + c

    In the polynomial 3x^2 + 4x + (-1-b)
    a = 3
    b = 4
    c = -1 - b
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Jul 2018
    From
    Canada
    Posts
    8

    Re: Linear Quadratic Systems Help

    Ok i get it now, so I can use a, b and c into the quadratic formula then?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Senior Member
    Joined
    Jan 2009
    Posts
    396
    Thanks
    67

    Re: Linear Quadratic Systems Help

    That's correct. I'll leave the rest to you.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Jul 2018
    From
    Canada
    Posts
    8

    Re: Linear Quadratic Systems Help

    Quote Originally Posted by MacstersUndead View Post
    That's correct. I'll leave the rest to you.
    Thank you! If you have the time would you mind checking my answer when i'm done?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Senior Member
    Joined
    Jan 2009
    Posts
    396
    Thanks
    67

    Re: Linear Quadratic Systems Help

    Sure.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Newbie
    Joined
    Jul 2018
    From
    Canada
    Posts
    8

    Re: Linear Quadratic Systems Help

    When i solve it i get

    x = - 2 +/- sqrt(7 + 3b) / 3 as my answer
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Senior Member
    Joined
    Jan 2009
    Posts
    396
    Thanks
    67

    Re: Linear Quadratic Systems Help

    That is the correct solution for x, but the question is asking for the restriction of b (the y-intercept).

    Recall the following.
    D, the determinant, is the expression under the square root. ie. b^2 - 4ac

    If D > 0, then there are two solutions.
    If D = 0, then there is a repeated solution
    If D < 0, then there are no real number solutions

    Since the question asks for the restriction on b when the parabola intersects the line in more than one place, D > 0
    D = 28 + 12b
    D > 0 -> D/4 > 0 -> 7 + 3b > 0

    Isolate b to determine the restriction on b.
    Thanks from fucc
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Newbie
    Joined
    Jul 2018
    From
    Canada
    Posts
    8

    Re: Linear Quadratic Systems Help

    Quote Originally Posted by MacstersUndead View Post
    That is the correct solution for x, but the question is asking for the restriction of b (the y-intercept).

    Recall the following.
    D, the determinant, is the expression under the square root. ie. b^2 - 4ac

    If D > 0, then there are two solutions.
    If D = 0, then there is a repeated solution
    If D < 0, then there are no real number solutions

    Since the question asks for the restriction on b when the parabola intersects the line in more than one place, D > 0
    D = 28 + 12b
    D > 0 -> D/4 > 0 -> 7 + 3b > 0

    Isolate b to determine the restriction on b.
    quick question can i use D = 7 + 3b or does it have to be D = 28 + 12b
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Senior Member
    Joined
    Jan 2009
    Posts
    396
    Thanks
    67

    Re: Linear Quadratic Systems Help

    By definition, D = 28 + 12b.
    However, you can divide D by a positive number as this will not change the sign of the inequality. The most common case you will find while studying quadratics is D/4. D/4 in this case is 7+3b.

    Starting at either D or D/4 doesn't change the solution for isolating b. In fact, starting at D, you'll divide by 4 on both sides anyway
    D = 28 + 12b > 0
    Dividing both sides by 4 yields D/4 = 7 + 3b > 0
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Newbie
    Joined
    Jul 2018
    From
    Canada
    Posts
    8

    Re: Linear Quadratic Systems Help

    As my final answer i got if b = 7/-3 then the line intersects with the parabola twice. since x can have 2 possible values there are 2 values of x that satisfies the equation
    Follow Math Help Forum on Facebook and Google+

  14. #14
    Senior Member
    Joined
    Jan 2009
    Posts
    396
    Thanks
    67

    Re: Linear Quadratic Systems Help

    Close. You need the inequality. Solving it yields b > -7/3. This means that if b > -7/3, then the parabola will intersect the line twice.
    Follow Math Help Forum on Facebook and Google+

  15. #15
    Newbie
    Joined
    Jul 2018
    From
    Canada
    Posts
    8

    Re: Linear Quadratic Systems Help

    My bad that's what I meant b > 7/-3. Also do I have to solve it for D = 0 and D > 0 as well or can I just stop at D < 0
    Follow Math Help Forum on Facebook and Google+

Page 1 of 2 12 LastLast

Similar Math Help Forum Discussions

  1. Quadratic-Quadratic Systems Algebraically Question
    Posted in the Pre-Calculus Forum
    Replies: 4
    Last Post: Dec 22nd 2012, 09:24 AM
  2. Replies: 0
    Last Post: Feb 13th 2011, 11:40 AM
  3. linear quadratic systems
    Posted in the Algebra Forum
    Replies: 3
    Last Post: Oct 29th 2010, 05:46 AM
  4. Modelling a quadratic - Systems of linear equations
    Posted in the Pre-Calculus Forum
    Replies: 30
    Last Post: May 12th 2010, 05:13 AM
  5. Replies: 7
    Last Post: Aug 30th 2009, 10:03 AM

Search Tags


/mathhelpforum @mathhelpforum