Need assistance with an unusual quadratic solution method

[gregi_2]

TL;DR

I have come across this strange method of solving degree 2 polynomials but I do not find the explanation provided to be very helpful.

I have come across this strange method of solving degree 2 polynomials but I do not find the explanation provided to be very helpful. Here is the method description:

"In the 16th century, mathematician Francois Viete solved quadratic equations by a unique substitution method. To solve an equation such as x^2 + 6x + 7 he substituted y + k for x, where k was a number to be determined. He then rearranged the equation as a quadratic in y and chose a value for k so that the coefficient of the linear term was zero. It was then an easy matter to solve for y, and then by adding the chosen value of k, the roots of the original equation were obtained."

I do not fully understand the steps described here and would like to see an example of this solution being applied to the equation x^2 + 6x + 7 = 0.

 

[anuttarasammyak]

x=y+k
(y+k)^2+6(y+k)+7=0
y^2+(2k+6)y+k^2+6k+7=0
Say 2k+6=0, k=-3
y^2=2
y=\pm \sqrt{2}
x=y+k=-3\pm \sqrt{2}

 

[hutchphd]

With the substitution $x=y+k$ the equation becomes $$(y+k)^2+6(y+k)+7=0$$ $$y^2+(6+2k)y+7+k^2+6k=0$$ choosing $k=-3$ gives $$y^2-2=0$$so $y=\pm\sqrt2$ and the solution is $$x=-3\pm\sqrt2$$
I think the quadratic formula is easier!

oops well its good we agree!

 

[anuttarasammyak]

I do not think this method is strange or unusual.
ax^2+bx+c=0
Applying this method
a(x+\frac{b}{2a})^2=\frac{b^2}{4a}-c=\frac{b^2-4ac}{4a}
x+\frac{b}{2a}=\pm\sqrt{\frac{b^2-4ac}{4a^2}}
x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}
We get usual formula.

EDIT And I do not find any other way to derive the formula.

 

[pasmith]

This is "making a linear change of variable to remove the linear term", ie. one of the standard methods of solving an arbitrary quadratic.

The other, "completing the square", is essentially the same, where we use the fact that (x + a)^2 = x^2 + 2ax + a^2 to make an intelligent guess as to what k is going to be.

Nowadays, of course, we don't teach these methods, but only the result of applying them to an arbitrary quadratic (ie. the standard formula).