Thread: Quadratic Roots
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Old December 3rd, 2008, 04:21 PM
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Hello, euclid2!

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

Quote:
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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