Please teach me how to solve this

[Mushini]

Please teach me how to solve this [now just a question]

58,000 = 200x^2 + 3800x


Is this a quadratic formula? A binomial?
I do apologize if this is in the wrong section.
I know that x = 10 is a solution, but I don't know how to solve this.

m = 29,000
i = 200
c = 2,000

m = x (ix/2 - i/2 + c)
29,000 = (x * (200x/2 - 200/2 + 2,000)
29,000 = (200x^2/2 + 1900x)
58,000 = 200x^2 + 3800x
200x^2 + 3800x - 58,000 = 0

I think I did that correctly. Why am I doing this? I'm trying to reverse this, to get x:
[(x -1) /2 *xi] + xc = m

c = s*(1 + 0.1*o)
i = s/10

s = 2000; i = 200 (these are constant)
o = 0
c = 2000

m = 2000 ; x = 1
m = 4200 ; x = 2
m = 6600 ; x = 3
m = 9200 ; x = 4
m = 12000 ; x = 5
m = 15000 ; x = 6
m = 18200 ; x = 7
m = 21600 ; x = 8
m = 25200 ; x = 9
m = 29000 ; x = 10, -29

Sorry - I know this isn't pretty or done right, and I hope I didn't make any errors.
So, I think this is a quadratic formula. Reading corresponding page on this site. I tried other sites but I was confused.
This is a quadratic formula! Thank you; the site's robot helped me out. Is this the only way [quadratic formula] to get x?

 

[JeffM]

58,000 = 200x^2 + 3800x


Is this a quadratic formula? A binomial?
This is a quadratic formula! Thank you; the site's robot helped me out. Is this the only way [quadratic formula] to get x?

Thank you for showing your work.

Technically, this is not the quadratic formula; it is a quadratic equation, and like all quadratic equations, it can be solved exactly using the quadratic formula.

There are at least six ways to solve quadratic equations, but some of them do not give exact answers, only approximations, and some of them are very time consuming. There are at least three ways of getting an exact answer, but one of them may be very time consuming.

(A) Factoring and the zero product property

\(\displaystyle 58000 = 200x^2 + 3800x \implies 290= x^2 + 19x.\) Factored out 200.

\(\displaystyle 290 = x^2 + 19x \implies x^2 + 19x - 290 = 0.\) Put into standard form.

\(\displaystyle x^2 + 19x - 290 = 0 \implies (x - 10)(x + 29) = 0.\) Factored the quadratic trinomial.

\(\displaystyle THUS\ x = 10\ or\ x = -29.\) Used the zero product property, which is:

If the product of two or more numbers is zero, at least one of the numbers being multiplied is zero

This factoring approach is very quick if you "see" a factoring. If you do not "see" one, it can take forever.

(B) Completing the square

This very general method always works, but it is usually quite cumbersome.

\(\displaystyle 58000 = 200x^2 + 3800x \implies 290= x^2 + 19x.\) Factored out 200.

\(\displaystyle 290 = x^2 + 19x \implies 290 + \left(\dfrac{19}{2}\right)^2 = x^2 + 19x + \left(\dfrac{19}{2}\right)^2.\)

Added the square of half the coefficient of x to both sides of the equation.

\(\displaystyle 290 + \left(\dfrac{19}{2}\right)^2 = x^2 + 19x + \left(\dfrac{19}{2}\right)^2 \implies \dfrac{4 * 290 + 19^2}{4} = \left(x + \dfrac{19}{2}\right)^2.\)

\(\displaystyle \dfrac{4 * 290 + 19^2}{4} = \left(x + \dfrac{19}{2}\right)^2 \implies x + \dfrac{19}{2} = \pm \sqrt{\dfrac{1521}{4}} = \pm \dfrac{39}{2} \implies\)

\(\displaystyle x = \dfrac{39}{2} - \dfrac{19}{2} = \dfrac{20}{2} = 10\ or\ x = -\dfrac{39}{2} - \dfrac{19}{2} = \dfrac{-58}{2} = - 29.\)

(C) Quadratic formula

\(\displaystyle 58000 = 200x^2 + 3800x \implies 290= x^2 + 19x.\) Factored out 200.

\(\displaystyle 290 = x^2 + 19x \implies x^2 + 19x - 290 = 0.\) Put into standard form.

\(\displaystyle x = \dfrac{-19 \pm \sqrt{19^2 - 4(1)(-290)}}{2 * 1} = \dfrac{-19 \pm \sqrt{361 + 1160}}{2} = \dfrac{-19 \pm \sqrt{1521}}{2}.\) Used quadratic formula.

\(\displaystyle x = \dfrac{-19 \pm \sqrt{1521}}{2} \implies x = \dfrac{-19 + 39}{2} = \dfrac{20}{2} = 10\ or\ x = \dfrac{-19 - 39}{2} = \dfrac{-58}{2} = -29.\)

Quadratic formula always gives you an exact answer.

 

[lookagain]

(A) Factoring and the zero product property

\(\displaystyle 58000 = 200x^2 + 3800x \implies 290= x^2 + 19x.\) Factored out 200.

\(\displaystyle 290 = x^2 + 19x \implies x^2 + 19x - 290 = 0.\) Put into standard form.

\(\displaystyle x^2 + 19x - 290 = 0 \implies > > (x - 10)(x - 29) < < = 0.\)

This is a typo. The second binomial factor is really (x + 29).

. . .

 

[JeffM]

Jeff, half hour in the corner...

Can I take my drink with me?

 

[MarkFL]

Denis is strict, but not unreasonable. :cool: