The sum of two numbers is 58

the sum of two numbers is 58.
those two numbers multiplied by each other give 814.

how would I go into solving this with the Quadratic formula?

Are you retarded?

no, why?

homework belongs on

You don't use the quadratic formula because there s no quatern.

you can get to it eventually tho, I'm looking for the way.

:(

Why though?

x+y = 58
x*y = 814

y = x-58

x (x-58)=814

x^2-58x-814 = 0

There's your quadratic.
Literally plug in to the quadratic formula.

(58+-(58^2 + 814*4)^.5)2

a + b = 56
ab = 814
a = 56 - b
(56 - b)b = 814
b^2 -56b + 814 = 0

a + b = 58
ab = 814

a = 58 - b

b(58 - b) = 814

-b^2 + 58b - 814 = 0

plug in those numbers into the formula.

multiplying the roots doesn't give the 814 as solution?

Factor 814.

2 * 412
2*2*206
2*2*103

103 is prime.
Therfore the roots are going to be irrational.
Best you can do is leave it in exact terms or round.

is it really 814 and not 841?

Those factors are wrong.

It's 2,11,37.

Still irrational though.

accidentally typed 59 instead of 58 and got

-x^2+59x-814=0
roots are exactly what I was looking for ( 22, 37)

how do I get to this position
-x^2+59x-814=0
with the info given?

(x-a)(x-b)=x^2-(a+b)x+ab

well friend 58 plus 58 equals hmm FIFTY-EIGHT!!!!!!!!!!!

a+b=59
a*b=814
59-b=814/b
b^2-59b+814=0
b=59/2+-sqrt((59/2)^2-814)
b=29.5+-7.5
b=37,22
a=814/b=22,37

actually

x>y
x+y=58
x*y + x =814

I'm confused now

General Form:
ax^2 + bx + c = 0

Roots:
x = 2, x = 2/3
(x - 2)( 6x- 4) = 0
6x^2 -16x +8 = 0

Product of roots = 4/3 = 8/6 = c/a
Sum of roots = 8/3 = 16/6 = b/a


So in your case 58 = b/a and 814 = c/a
Assume a = 1

x^2 + 58x + 814 = 0

x = -29 +- 3sqrt(3)

this doesn't work, it's supposed to be -x^2 +59 - 814 =0
but how did the 59 get here?

Oh yah for a quadratic sum of roots = -b/a silly me

>plebs using the quadratic formula instead of using the perfect square formula and deriving the roots through actual math instead of memorization

elaborate?

honestly op should get a public ban for this thread, shitty mods don't care

Example

a^2+6a+8=0
a^2+6a+8+1=1
(a+3)^2 = 1
(a+3)^2 = 1
a+3 = plus or minus sqrt(1)
a = -2, a = -4

basically you divide the second term by two and then square it to find the number you need for (a+k)^2 to work and then solve for a, aka the perfect square formula.


protip for plebs, memorization of long ass formulas is CC engineer tier.

fuck off, doesn't contribute and shits on others, nice one.

you need to kill yourself asap

solve the question and I will.

>here is my homework.
>how would you solve my homework?

nope. sry.

a+b = summation
a*b = multiplied
a = 0.5*(summation-sqrt(pow(summation,2)-(4*multiplied)))
b = 0.5*(sqrt(pow(summation,2)-(4*multiplied))+summation)

You learned me something new today?

wrong
correct

Still don't know what I'm supposed to do.

814=x(58-x)+x
try now

This thread is some high level shit

Divide 814 by 58

OP, did you accidentally type 814 instead of 841 ([math]\frac{58}{2}^2[/math]) or did you do that in purpose for people to solve?

I think he mistyped 58 with 59
x^2-59x+814=0
roots correspond with 814.

x=(y*(summation-y))+y
0.5*(sqrt((-4*multiplied)+pow(summation,2)+(2*summation)+1)+summation+1)

...

why the plus x?! it just says: "the two numbers multiplied with each other" aka "a*b"
-----------
[math]a+b=58 \leftrightarrow a=58-b[/math]
[math]a \cdot b=814 \leftrightarrow a=\frac{814}{b}[/math]

[math](58-b)\cdot b=814 \leftrightarrow b^2 -58b+814=0[/math]
[math]\leftrightarrow b_{1,2}=\frac{58}{2}\pm \sqrt{(\frac{58}{2})^2-814}[/math]
[math]\rightarrow b_1 = \frac{58}{2}+ \sqrt{(\frac{58}{2})^2-814} = 29+3 \cdot \sqrt{3}\approx 34,2[/math]
[math]\rightarrow b_2 = \frac{58}{2}- \sqrt{(\frac{58}{2})^2-814 =29-3 \cdot \sqrt{3} \approx 23,8 [/math]
[math]a=58-29-3 \cdot \sqrt{3} = b_2 [/math]
[math]a \cdot b = 814 [/math]
but b values back into the quadratic formulas and they do give zero. i don't see a problem here.

and i don't know where the fuck that extra "+ x" in is supposed to be.

...

>mfw this thread

to get
-x2+59x-814=0

you gotta have 814=x(58-x)+x

I don't know what's the purpose of the thread anymore.

...