The WORST way of solving a quadratic equation

Have you ever considered using the rational zero theorem (rational root theorem) with the synthetic division to solve a regular quadratic equation? If not, don't miss out on this math video!
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[jdominoes]

Nah this isn't the worst, the worst is literally just plugging in random numbers to the equation until it's correct

[SeegalMasterPlayz]

Oh, and to make this even worse we can divide the quadratic equation itself by the factor rather using synthetic division.

[ВладиславЗиновей]

The sum of the coefficients is zero. For the expression to be zero, x must be equal to 1. According to the Vieta's formula, the second root of the equation is -11/6.

[joshuahillerup]

I think checking all 16 possible rational zeroes would be even worse, instead of doing synthetic division

[Misteribel]

My way of doing it is to spot instantly that 5+6 equals 11. The first solution is 1, and then clearly, the multiplication factor to get 6 for the square is 6, which gives us (x - 1)(6x + 11). There is no need for synthetic division, quadratic formula, or rational zeroes.

[arantheo]

"There's more than one way to do things!"
Interestingly, the word coefficient, which in Latin means "cooperating to produce a result", seems to fit nicely. Note polynomial comes from polys (Greek for "many") and nomen (Latin for "name").
Rational Zero Theorem
the possible rational zeros of a polynomial function have the form p/q where p is a factor of the constant term and q is a factor of the leading coefficient

[GaryBricaultLive]

An easier way to to first multiply the C term by the A term, 6 * -11 yielding
(x^2) + 5x - 66 = 0 now factor out the binomials of the new quadratic
( x - 6)( x + 11) = 0 now divide the 6 back out of the two integers
(x - 6/6)(x + 11/6) = 0
(x - 1)(x + 11/6) = 0 now solve for x
x - 1 = 0
x + 11/6 = 0 giving the two results
x = 1 and X = -11/6
I learned this trick when there is a multiplier associated with the X^2 term

[Jokir]

Change to x^2 +5x-66=0 and divide final zeroes by 6. (x+11)(x-6)=0. x=-11/6 or x=1

[briant]

Factoring is easy.
6x² + 5x - 11
Factors of 11 are +/-1 and -/+11
11 - 6 = 5, so 1x and 6x
Then what arrangement fits?
(x - 1) (6x + 11)
So x = 1 or -11/6

[officialfxlxm]

What are your thoughts on this method?
6x²+5x-11=0
Divide everybody by 6
x²+ 5/6x -11/6 = 0
Assume solutions are a± √(b)
-2a= 5/6 a²-b= -11/6
a=-5/12 (-5/12)² - b = -11/6
25/144+11/6=b
b=289/144
So, solutions are -5/12 ± √(289/144)
Which is x=1 and x=-11/6 when you solve.

[ProactiveYellow]

6x²+5x-11=0 multiply by the leading coefficient
(6x)²+5(6x)-66=0 monic quadratic in 6x, odd linear coefficient, so multiply by 2²=4
(12x)²+10(12x)-264=0 Even middle term for integer division later
(12x)²+10(12x)=264 LHS prepped to complete the square
(12x)²+10(12x)+5²=25+264 square completed
(12x)²+5(12x)+5(12x)+5²=289 separate middle term for factoring, simplify RHS
(12x)(12x+5)+5(12x+5)=289 factor by grouping
(12x+5)²=289=17² factor as far as possible
add product of roots to both sides
(12x+5)(12x+22)=17(12x+22) factor both sides
(12x+22)(12x+5)-(12x+22)17=0 move to LHS
(12x+22)(12x-12)=0 factor and reduce
(6x+11)(x-1)=0 divide by 24=2×12 in appropriate factors.
(6x+11)=0 or (x-1)=0 by zero product property of the real numbers
x=-11/6 or x=1 solving linear equations.

The most drawn out, full scale form of both the "slide and divide" method into "completing the square, " followed by full factoring of a difference of squares, all while in the integers when possible. This is the "best way" with all the steps shown.

[DefenderTerrarian]

For Quadratic, form of ax^2+bx+c, find the factors of ac, call them d, e. The stipulation is that d+e=b. Once you have found suitable d and e. Rewrite the equation as ax^2+dx+ex+c, or conversely, swap dx+ex. Consider the two halves of the equation, ax^2+dx, and ex+c. Pull out common factors. If you've done it right, you should end up with fx(gx+h)+i(gx+h), where all letters are variables, cut us some slack, We're on mobile, then factor out the (gx+h), and you're left with (fx+i)(gx+h), which are the factors of ax^2+bx+c.
It sounds a lot more complicated when explained with variables, but is actually rather intuitive.

[kausarali]

It's not the worst way of solving but more innovative way to Express quadratic equation in terms of synthetic division method

[aymanelaidi-tn]

can you make a video on the basics of of derivatives and integrations pls?

[undefinedperson]

A simple way is to find obvious zeroes first. Try 1, -1, 2, -2, 3, -3.

[Imotbro]

I once tried to solve a quadratic using a system of equations as factoring



It stuck me in a infinite loop.

[solidpixel]

Once you get the first factor is (x - 1) it's easy to see what the second factor has to be without doing the synthetic division. The first part of the second factor has to be 6 to get 6x^2 and the second part has to be 11. So it's (6x + 11)

[vikramrawat]

There are many ways to solve quadratic equations
I know all of them
But i don't have enough space to write them down

[barneyDcaller]

6x²+5x-11 = 0
x²+5x-66 = 0
(x+11)(x-6) = 0
(x+11/6)(x-6/6) = 0
(6x+11)(x-1) = 0
6x+11 = 0 ; x = -11/6
x-1 = 0 ; x = 1

[michaelsorice]

We can increase the efficiency somewhat by using synthetic evaluation rather than direct evaluation. That way, we get both the verification that a 0 has been found and the corresponding quotient polynomial in one step.