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Thread: Quadratic Roots

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    Senior Member euclid2's Avatar
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    Quadratic Roots

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

    $\displaystyle f(x) = k(x - r_1)(x - r_2)$

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

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

    $\displaystyle f(x) = k(x - r_1)(x - r_2)$

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

    now, substitute $\displaystyle r_1 = 2-\sqrt{3}$ and $\displaystyle r_2 = 2 + \sqrt{3}$ to determine the basic quadratic in [...], then use the fact that $\displaystyle f(1) = -2$ to find $\displaystyle k$
    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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    Hello, euclid2!

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


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

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

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

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


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

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


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

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

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

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


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

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


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

    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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    $\displaystyle f(x) = k(x - r_1)(x - r_2)$

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

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

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

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

    since $\displaystyle f(1) = -2$

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

    $\displaystyle -2 = k(-2)$

    $\displaystyle k = 1$

    so ...

    $\displaystyle f(x) = x^2 - 4x + 1$
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