How to solve this quadratic equation?

please can you help me solve this quadratic equation as i am a but stuck.

please show your working.

3x^2-y^2=3 2x-1=y

6 Answers

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  • 7 years ago
    Best Answer

    3x^2 - y^2 = 3

    2x - 1 = y

    3x^2 - (2x - 1)^2 = 3

    3x^2 - (4x^2 - 4x + 1) = 3

    3x^2 - 4x^2 + 4x - 1 = 3

    -x^2 + 4x - 4 = 0

    x^2 - 4x + 4 = 0

    (x - 2)^2 = 0

    x = 2

    and if x = 2, then y = 2(2) - 1 = 3

    solution: (x , y) = (2 , 3)

    edit: interpreting this geometrically:

    the first equation is a hyperbola

    the second equation is a line

    so there is one point of intersection, and the line is a tangent of the hyperbola

    check:

    3(2^2) - (3^2) = 12 - 9 = 3 yep

    2(2) - 1 = 3 yep

    (2 , 3)

  • 7 years ago

    3x²-y²=3 (1)

    2x-1=y (2)

    Let's put y in function of x in (1).

    3x²-(2x-1)²=3

    3x²-(4x²-4x+1)=3

    -x²+4x-4=0

    x²-4x+4=0

    (x-2)²=0

    x-2=0

    x=2

    y=2x-1

    y=2.2-1

    y=3

    The system of equations has got an unique solution.

    x=2

    y=3

  • Tony
    Lv 7
    7 years ago

    3x^2 - y^2 = 3, 2x - 1 = y. Therefore 3x^2 - (2x - 1)^2 = 3, or 0 = x^2 - 4x + 4 = (x - 2)^2, so x = 2. Since 2x - 1 = y we have 2(2) - 1 = y, or y = 3.

  • Como
    Lv 7
    7 years ago

    3x² - (2x - 1)² = 3

    3x² - (4x² - 4x + 1) = 3

    3x² - 4x² + 4x - 1 = 3

    x² - 4x + 4 = 0

    (x - 2)(x - 2) = 0

    x = 2

    y = 3

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  • Anonymous
    7 years ago

    1. Substitute eqn. 2 for y in eqn.1

    2. Expand the the new y^2 term

    3. Regroup and cancel so in form ax^2+bx+c=0

    4. Use the standard quadratic equation formula.

  • 3 years ago

    i exploit the "new and more suitable factoring AC approach" (Google or Yahoo seek) to unravel the equation 2y^3 + 7y^2 - 30y = 0. First factor out y: y(2y^2 + 7y - 30) = 0. next, resolve the quadratic equation in parentheses. the two roots have opposite indications. Compose the aspects of ac = -60 with first numbers destructive. they are: (-a million, 60)(-2, 30)(-3, 20)(-4, 15)(-5, 12). end. This final sum is -5 +12 = 7 = b. next, replace in the equation the term (7y) by capacity of the two words (-5y) and (12y), then put in elementary factor: 2y^2 - 5y + 12y - 30 = 2y(y + 6) - 5(y + 6) = (y + 6)(2y - 5) = 0. next, resolve the binomials: y + 6 = 0 --> y = -6 2y - 5 = 0 --> y = 5/2

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