Determine the equation of the quadratic function that passes through $(2,5)$ if the roots of the corresponding quadratic equation are $1+\sqrt 5$ and $1-\sqrt 5$.
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1Hint: The quadratic function has shape $k(x^2-2x-4)$. Remark: Your post would be much improved if you indicated what you had tried. – André Nicolas Jun 20 '16 at 00:17
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hello sir, what do you mean by that – IAmAboutToFailLol Jun 20 '16 at 00:18
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I found a quadratic whose roots are the given ones. The easiest way to do this is to recall that if the quadratic $x^2+bx+c$ has roots $r$ and $s$, then $b=-(r+s)$ and $c=rs$. Now please play with it for a while. – André Nicolas Jun 20 '16 at 00:21
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One of the results of the fundamental theorem of algebra is that every degree $n$ polynomial may be written in the form $f(x)=k(x-\alpha_1)(x-\alpha_2)\cdots(x-\alpha_n)$ where $\alpha_1,\dots,\alpha_n$ are the roots (possibly complex and possibly repeated).
You are told that you are working with a quadratic and that its two roots are $(1+\sqrt{5})$ and $(1-\sqrt{5})$. You are also told that $f(2)=5$ as it passes through the point $(2,5)$.
These imply the following two things:
$f(x)=k(x-1-\sqrt{5})(x-1+\sqrt{5})$ and $f(2)=5$
Solving for $k$...
we have $5=k(2-1-\sqrt{5})(2-1+\sqrt{5})=k(1-\sqrt{5})(1+\sqrt{5})=k(1-5)=-4k$
Implying that
$k=-\frac{5}{4}$
Thus, your quadratic is:
$f(x)=-\frac{5}{4}(x-1-\sqrt{5})(x-1+\sqrt{5})$
which can simplify if desired as
$f(x)=-\frac{5}{4}x^2+\frac{5}{2}x+5$
JMoravitz
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