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Math Help - Quadratic Roots

  1. #1
    Senior Member euclid2's Avatar
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    Quadratic Roots

    Find the equation of the quadratic given it's real roots 2-\sqrt 3 and 2 + \sqrt3 which passes through the point (1,-2)
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  2. #2
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    if r_1 and r_2 are two roots of a quadratic ...

    f(x) = k(x - r_1)(x - r_2)

    f(x) = k[x^2 -(r_1 + r_2)x + r_1 \cdot r_2]

    now, substitute r_1 = 2-\sqrt{3} and r_2 = 2 + \sqrt{3} to determine the basic quadratic in [...], then use the fact that f(1) = -2 to find k
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  3. #3
    Senior Member euclid2's Avatar
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    Quote Originally Posted by skeeter View Post
    if r_1 and r_2 are two roots of a quadratic ...

    f(x) = k(x - r_1)(x - r_2)

    f(x) = k[x^2 -(r_1 + r_2)x + r_1 \cdot r_2]

    now, substitute r_1 = 2-\sqrt{3} and r_2 = 2 + \sqrt{3} to determine the basic quadratic in [...], then use the fact that f(1) = -2 to find k
    I'm not sure exactly what you mean, can you please work it down further? I have never learned this formula.
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  4. #4
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    Hello, euclid2!

    There are several approaches to this problem.
    . . Here's one of them . . .


    Find the equation of the quadratic given its real roots 2-\sqrt 3 and 2 + \sqrt3
    which passes through the point (1,-2)

    If the quadratic, ax^2 + bx + c, has roots p\text{ and }q, then: . p+q \:=\:-\frac{b}{a},\;\;pq \:=\:\frac{c}{a}

    So we have: . \begin{array}{ccc}p+q &=&4 \\ pq &=& 1\end{array}\quad\Rightarrow\quad\begin{array}{ccc  }\frac{b}{a} &=& \text{-}4 \\ \frac{c}{a} &=& 1 \end{array} . \Rightarrow\quad\begin{array}{ccc}b &=&\text{-}4a \\ c &=&a \end{array}

    . . The function (so far) is: . f(x) \:=\:ax^2 -4ax + a


    Since (1,-2) is on the graph: . a\!\cdot\!1^2 - 4a\!\cdot\!1 + a \:=\:-2 \quad\Rightarrow\quad -2a \:=\:-2

    . . Hence: . a\:=\:1,\;\;b\:=\:-4,\;\;c\:=\:1


    Therefore: . f(x)\;=\;x^2 - 4x + 1

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  5. #5
    Senior Member euclid2's Avatar
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    Quote Originally Posted by Soroban View Post
    Hello, euclid2!

    There are several approaches to this problem.
    . . Here's one of them . . .



    If the quadratic, ax^2 + bx + c, has roots p\text{ and }q, then: . p+q \:=\:-\frac{b}{a},\;\;pq \:=\:\frac{c}{a}

    So we have: . \begin{array}{ccc}p+q &=&4 \\ pq &=& 1\end{array}\quad\Rightarrow\quad\begin{array}{ccc  }\frac{b}{a} &=& \text{-}4 \\ \frac{c}{a} &=& 1 \end{array} . \Rightarrow\quad\begin{array}{ccc}b &=&\text{-}4a \\ c &=&a \end{array}

    . . The function (so far) is: . f(x) \:=\:ax^2 -4ax + a


    Since (1,-2) is on the graph: . a\!\cdot\!1^2 - 4a\!\cdot\!1 + a \:=\:-2 \quad\Rightarrow\quad -2a \:=\:-2

    . . Hence: . a\:=\:1,\;\;b\:=\:-4,\;\;c\:=\:1


    Therefore: . f(x)\;=\;x^2 - 4x + 1

    Thank you very much, although can you edit this to show me where you are getting those numbers from? Thanks, again.
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  6. #6
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    f(x) = k(x - r_1)(x - r_2)

    f(x) = k[x^2 -(r_1 + r_2)x + r_1 \cdot r_2]

    r_1 + r_2 = (2 - \sqrt{3}) + (2 + \sqrt{3}) = 4

    r_1 \cdot r_2 = (2 - \sqrt{3})(2 + \sqrt{3}) = 4 - 3 = 1

    f(x) = k(x^2 - 4x + 1)

    since f(1) = -2

    -2 = k[(1)^2 - 4(1) + 1]

    -2 = k(-2)

    k = 1

    so ...

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