What is the equation of the quadratic function through $(2,5)$ with roots $1+\sqrt 5$ and $1-\sqrt 5$?

Question

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 $1-\sqrt 5$.

Answer 1

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

$f(x)=-\frac{5}{4}x^2+\frac{5}{2}x+5$