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Math Help - Quadratic Equation

  1. #1
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    Quadratic Equation

    Write the quadratic equation in standard form. Then solve using the quadratic formula.

    2 - 3x + x^2 = 0

    So in standard form, this is what I got:

    x^2 + 3x - 2 = 0

    How do I solve it using the quadratic formula?
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  2. #2
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    Quote Originally Posted by SarynJumail View Post
    Write the quadratic equation in standard form. Then solve using the quadratic formula.

    2 - 3x + x^2 = 0

    So in standard form, this is what I got:

    x^2 + 3x - 2 = 0

    How do I solve it using the quadratic formula?
    -b+- SQRT((b^2)-4ac)
    so..

    -3 +- SQRT((3^2)-4(1)(-2))/(2)(1)
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  3. #3
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    -3 +- SQRT(9) + (8) / (2)

    -3 +- SQRT (17) / (2)

    So where do I go from here?
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  4. #4
    Member ~berserk's Avatar
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    Quote Originally Posted by SarynJumail View Post
    -3 +- SQRT(9) + (8) / (2)

    -3 +- SQRT (17) / (2)

    So where do I go from here?
    That should be it since sqrt(17) is not a perfect square so you get -3√(17)/2, because you cannot cancel anything out, as your answer. Just by looking at the discriminate , in this case 17, you can determine that the roots of your equation will be real,unequal and irrational, also it has two x-intercepts if you are graphing it.( For future reference)
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  5. #5
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by SarynJumail View Post
    Write the quadratic equation in standard form. Then solve using the quadratic formula.

    2 - 3x + x^2 = 0

    So in standard form, this is what I got:

    x^2 + 3x - 2 = 0

    How do I solve it using the quadratic formula?
    The quadratic formula statest that : x=\frac{-b\pm \sqrt{b^2-4ac}}{2a} where a is the coefficient of x^2, b is the coefficient of x and c is the last term.

    Plugging into the quadratic formula : x=\frac{-3\pm \sqrt{(-3)^2-4(1)(-2)}}{2}\implies x=\frac{-3\pm\sqrt{17}}{2}.
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  6. #6
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    Okay.
    So -3 + SQRT of 7 / 2 and -3 - SQRT of 7/2.
    That's right?
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  7. #7
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by SarynJumail View Post
    Okay.
    So -3 + SQRT of 7 / 2 and -3 - SQRT of 7/2.
    That's right?
    \frac{-3\pm \sqrt{17}}{2} = \frac{-3+\sqrt{17}}{2} \ or \ \frac{-3-\sqrt{17}}{2}.

    I hope that this is clear.
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  8. #8
    Rhymes with Orange Chris L T521's Avatar
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    Derivation

    Derivation of the quadratic formula:

    ax^2+bx+c=0

    Subtract both sides by c.

    ax^2+bx=-c

    Divide through by a.

    x^2+\frac{b}{a}x=-\frac{c}{a}

    Complete the square on the left side (to keep equation balanced, add the new term to the right side as well)

    x^2+\frac{b}{a}x+\left(\frac{b}{2a}\right)^2=\frac  {b^2}{4a^2}-\frac{c}{a}

    The left side becomes a perfect square. Combine the terms on the right side of the equation:

    \left(x+\frac{b}{2a}\right)^2=\frac{b^2}{4a^2}-\frac{c}{a}\cdot\frac{(4a)}{(4a)}\implies \left(x+\frac{b}{2a}\right)^2=\frac{b^2-4ac}{4a^2}

    Take the square root of both sides:

    x+\frac{b}{2a}=\pm\sqrt{\frac{b^2-4ac}{4a^2}}
    \implies x+\frac{b}{2a}=\pm\frac{\sqrt{b^2-4ac}}{2a}

    Solve for x:

    x=-\frac{b}{2a}\pm\frac{\sqrt{b^2-4ac}}{2a}

    \color{red}\boxed{\therefore x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}}
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