The Poh-Shen Loh method is another method for solving quadratic equations, the method was pioneered by Prof. Poh Shen-Loh, a professor of mathematics at the Carnegie Mellon University. Although the method was invented by ancient Babylonians, Poh Shen-Loh took it upon himself to refurbish and upgrade the method. The method although looks like an extension of the completing the square method and has been tested to work for all conditions of quadratics.
Solve $x^2-6x+8=0$ using the Poh Shen-Loh method.
We are going to factor the equation into
$x^2-6x+8=(x-x_1)(x-x_2)$
$x^2-(x_1+x_2)x+x_1x_2$
Where:
Sum: $x_1+x_2=6$
Product: $x_1x_2=8$
Now we obtain the midpoint from the sum: $\frac{x_1+x_2}{2}=\frac{6}{2}=3$
Let $u$ be the distance between $x_1$ and $3$ and between $3$ and $x_2$.
Let $x_1=3-u$ and $x_2=3+u$
And $x_1x_2=8\Rightarrow(3-u)(3+u)=8$
$9-u^2=8\Rightarrow{u^2=9-1=1}$
Now that we have obtained the value of $u=1$, substitute $u$ into $x_1$ and $x_2$.
$x_1=3-u=3-1=2$ and $x_2=3+u=3+1=4$.
Hence our root is $x_{1,2}=2,4$.
Now let's take another example with repeated roots.
Solve $x^2+4x+4=0$ using the Poh Shen-Loh method.
We rewrite the equation into $x^2-(-4x)+4=0$
Because we need our equation to be in the form $x^2-(x_1+x_2)x+x_1x_2=0$.
Now, the sum=$x_1+x_2=-4$ and midpoint=$\frac{x_1+x_2}{2}=\frac{-4}{2}=-2$ and the product=$x_1x_2=4$.
Let $u$ be the distance between $x_1$ and $-2$ and between $-2$ and $x_2$ then $x_1=-2-u$ and $x_2=-2+u$.
$x_1x_2=4\Rightarrow(-2-u)(-2+u)=4$
$4-u^2=4$
$u^2=0$
$u=0$
Subsitute $u$ into $x_1$ and $x_2$.
$x_1=-2-u=-2-0=-2$
$x_2=-2+u=-2+0=-2$
Therefore, the roots of the equation is $x_{1,2}=-2,-2$
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