Need assistance with an unusual quadratic solution method

In summary, In the 16th century, Francois Viete invented a unique substitution method for solving quadratic equations by substituting y+k for x and rearranging the equation to solve for y, then adding the chosen value of k to obtain the roots of the original equation. This method can be applied to any quadratic equation, but nowadays, the more commonly taught method is the quadratic formula.
  • #1
gregi_2
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TL;DR Summary
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.
 
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  • #2
[tex]x=y+k[/tex]
[tex](y+k)^2+6(y+k)+7=0[/tex]
[tex]y^2+(2k+6)y+k^2+6k+7=0[/tex]
Say 2k+6=0, k=-3
[tex]y^2=2[/tex]
[tex]y=\pm \sqrt{2}[/tex]
[tex]x=y+k=-3\pm \sqrt{2}[/tex]
 
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  • #3
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!
 
  • #4
I do not think this method is strange or unusual.
[tex]ax^2+bx+c=0[/tex]
Applying this method
[tex]a(x+\frac{b}{2a})^2=\frac{b^2}{4a}-c=\frac{b^2-4ac}{4a}[/tex]
[tex] x+\frac{b}{2a}=\pm\sqrt{\frac{b^2-4ac}{4a^2}}[/tex]
[tex]x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]
We get usual formula.

EDIT And I do not find any other way to derive the formula.
 
Last edited:
  • #5
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 [itex](x + a)^2 = x^2 + 2ax + a^2[/itex] to make an intelligent guess as to what [itex]k[/itex] 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).
 

FAQ: Need assistance with an unusual quadratic solution method

What is a quadratic equation?

A quadratic equation is an algebraic equation in the form of ax^2 + bx + c = 0, where a, b, and c are constants and x is the variable. It can have two solutions or roots, which can be found using various methods.

What is an unusual quadratic solution method?

An unusual quadratic solution method is a non-traditional approach to solving quadratic equations. It may involve using unique formulas or techniques that are not commonly taught in traditional math courses.

Why would I need assistance with an unusual quadratic solution method?

Sometimes, traditional methods of solving quadratic equations may not work for certain equations. In these cases, an unusual quadratic solution method may be necessary to find the solution. It can also be helpful for challenging or complex equations.

Are there any benefits to using an unusual quadratic solution method?

Yes, using an unusual quadratic solution method can provide a different perspective and approach to solving equations. It can also help develop critical thinking and problem-solving skills.

Where can I find assistance with an unusual quadratic solution method?

You can find assistance with an unusual quadratic solution method through online resources, math tutors, or by consulting with other mathematicians or scientists. It may also be helpful to explore different mathematical forums or communities for tips and advice.

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