November 25th, 2020 at 1:58:58 PM
permalink

I ran into this article about a mathematician had found easier way to solve quadratic equations. Here is the link to that article:

https://www.popularmechanics.com/science/math/a30152083/solve-quadratic-equations/?utm_source=facebook&utm_campaign=socialflowFBPOP&utm_medium=social-media&fbclid=IwAR05kBsmSQoYzpmQ7zfBEaCAzQCsl5ipAI9M00VxozFIneyxDu9mcatWce8

I have this equation in the form of ax^2 + bx + c =0, where a = 1, b = -131, and c = 3870, and attempted to solve it using the new method, but wasn't successful. Is this equation quadratic?

https://www.popularmechanics.com/science/math/a30152083/solve-quadratic-equations/?utm_source=facebook&utm_campaign=socialflowFBPOP&utm_medium=social-media&fbclid=IwAR05kBsmSQoYzpmQ7zfBEaCAzQCsl5ipAI9M00VxozFIneyxDu9mcatWce8

I have this equation in the form of ax^2 + bx + c =0, where a = 1, b = -131, and c = 3870, and attempted to solve it using the new method, but wasn't successful. Is this equation quadratic?

November 25th, 2020 at 2:43:31 PM
permalink

Quote:777I ran into this article about a mathematician had found easier way to solve quadratic equations. Here is the link to that article:

https://www.popularmechanics.com/science/math/a30152083/solve-quadratic-equations/?utm_source=facebook&utm_campaign=socialflowFBPOP&utm_medium=social-media&fbclid=IwAR05kBsmSQoYzpmQ7zfBEaCAzQCsl5ipAI9M00VxozFIneyxDu9mcatWce8

I have this equation in the form of ax^2 + bx + c =0, where a = 1, b = -131, and c = 3870, and attempted to solve it using the new method, but wasn't successful. Is this equation quadratic?

Yes, x

^{2}-131x+3870=0 is quadratic.

I like that new method, and it worked for me on your equation.

The sum of the answers should be 131, which means they are in the form 131/2 - u and 131/2 + u. That is x = 65.5 - u and x = 65.5 + u.

The product of answers equals 3870:

(65.5-u)(65.5+u) = 65.5

^{2}- u

^{2}= 3870

Therefore u

^{2}= 4290.25 - 3870 = 420.25.

So, u = (+ or -)sqrt(420.25) = (+ or -)20.5.

Then x - u = 45, and x + u = 86.

And both answers check.

Thanks for the link!

November 25th, 2020 at 2:46:50 PM
permalink

Quote:777I have this equation in the form of ax^2 + bx + c =0, where a = 1, b = -131, and c = 3870, and attempted to solve it using the new method, but wasn't successful. Is this equation quadratic?

Yes, it is, and there's nothing really new about the method.

x^2 - 131 x + 3870 = 0 has the sum of its roots 131 and the product of its roots 3870.

The roots are 131/2 + z and 131/2 - z.

The product is (131/2)^2 - z^2= 3870

This is (131/2)^2 - 3870 = z^2

z^2 = (17,161 - 15,480) / 4 = 1681 / 4

z = 41 / 2

A similar way of doing this is called "completing the square": this stems from (x - a)^2 = x^2 - 2a x + a^2.

Start with x^2 - 131 x + 3870 = 0

Subtract 3870 from both sides: x^2 - 131 x = -3870

Add (-131/2)^2 to both sides: x^2 - 131 x + (-131/2)^2 = (-131/2)^2 - 3870

The left side is (x - 131/2)^2; the right side is 17,161 / 4 - 15,480 / 4 = 1681 / 4 = (41 / 2)^2

Take the square root of both sides, keeping in mind that the square root of X^2 is both X and -X:

+/- (x - 131/2) = 41 / 2

x = 131 / 2 +/- 41 / 2 = 86, 45

(x - 131/2) = +/- sqrt(2041) / 2

x = 131/2 +/- sqrt(2041) / 2

November 25th, 2020 at 3:05:47 PM
permalink

Your equation is quadratic. I get the same roots (86, 45) using the new method as the traditional quadratic formula.

I rather like the "new" method and feels it offers some insight in developing the original algebra skills of factoring quadratics (with nice answers) by guessing at roots and seeing which work. For actual determination of the roots, it is is literally just doing the two parts of quadratic equation separately.....where a=1

average = -b/(2a) = -b/2

and solving for

u= (+ or -) sqrt(b^2 - 4ac) / (2a)

= (+ or -) sqrt(b^2 - 4c) /2

= (+ or -) sqrt[ (b/2)^2 - c)]

roots are (average + u) and (average - u)

I think the manipulation of the steps solving for "u" is no easier for irrational or complex roots than plugging into the quadratic formula.

average = -b/2 = -(-131)/2 = 131/2

(131/2 + u)*(131/2 - u) = 3870

(131^2)/4 - u^2 = 3870

u^2 = (131^2)/4 - 3870 = 420.25

u=20.5

roots are

131/2+20.5 = 65.5+20.5 = 86

131/2-20.5 = 65.5-20.5 = 45

I rather like the "new" method and feels it offers some insight in developing the original algebra skills of factoring quadratics (with nice answers) by guessing at roots and seeing which work. For actual determination of the roots, it is is literally just doing the two parts of quadratic equation separately.....where a=1

average = -b/(2a) = -b/2

and solving for

u= (+ or -) sqrt(b^2 - 4ac) / (2a)

= (+ or -) sqrt(b^2 - 4c) /2

= (+ or -) sqrt[ (b/2)^2 - c)]

roots are (average + u) and (average - u)

I think the manipulation of the steps solving for "u" is no easier for irrational or complex roots than plugging into the quadratic formula.

average = -b/2 = -(-131)/2 = 131/2

(131/2 + u)*(131/2 - u) = 3870

(131^2)/4 - u^2 = 3870

u^2 = (131^2)/4 - 3870 = 420.25

u=20.5

roots are

131/2+20.5 = 65.5+20.5 = 86

131/2-20.5 = 65.5-20.5 = 45