Quadratic Roots

euclid2 · #1 $Old$ December 3rd, 2008, 03:26 PM

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

skeeter · #2 $Old$ December 3rd, 2008, 04:02 PM

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 $k$

euclid2 · #3 $Old$ December 3rd, 2008, 04:16 PM

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.

Soroban · #4 $Old$ December 3rd, 2008, 04:21 PM

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

euclid2 · #5 $Old$ December 3rd, 2008, 04:29 PM

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.

skeeter · #6 $Old$ December 3rd, 2008, 04:56 PM

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

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

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

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

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

since $f(1) = -2$

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

$-2 = k(-2)$

$k = 1$

so ...

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