Question

What are the steps to solving a quadratic equation? @dan815 @Clalgee

Answer 1

First, make the numerical coefficient of the x/2 term equal to 1.

Than you should rewrite the equation with the constant by itself on the right side of the equation.

Next, take half of the numerical coefficient of the x term, square it, and add this quantity to both sides of the equation.

Factor the trinomial into the square of a binomial.

Use the square root property to take the square root of both sides of the equation.

Your only option is to solve for x.

Be sure to check the solution in the original equation.

Answer 2

@dan815 @Clalgee @ganeshie8

Answer 3

@dan815 @Clalgee @ganeshie8

Answer 4

maybe you are dealing with a specific problem ?

Answer 5

COuld you help?

Answer 6

depends.... I am very poor at math.

Answer 7

No.. You are an Honorary Professor of Math. I doubt that.

Answer 8

you want the general steps, or a specific problem. If a general order of operation, then do you want to complete the square or use the quadratic formula?

Answer 9

Are the steps that @Clalgee said above correct? I just want correct and accurate steps in order to complete the problems I am having difficult with.

Answer 10

my math ability is debatable, and any judgement about it would be subjective. I don't want to discuss that, that is off-topic.

And yes, Clalgee is correct. She is showing how to complete the square with any quadratic equation.

Answer 11

Factoring is not always an option: (unless the discriminant is a perfect square)

\(\scriptsize\color{ slate }{\scriptsize{\bbox[5pt, royalblue ,border:2px solid royalblue ]{~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ }}}\)

Quadratic formula.

\(\normalsize\color{ slate }{\Huge{\bbox[5pt, lightcyan ,border:2px solid black ]{ \LARGE{x=~} \huge{ \frac{-\color{magenta}{b} \pm\sqrt{ \color{magenta}{b} ^2-4 \color{blue}{a} \color{red}{c}}}{2 \color{blue}{a}} }~ }}}\)

when the equation is \(\LARGE\color{black}{ \color{blue}{a} x^2+ \color{magenta}{b}x+ \color{red}{c}=0 }\).

Answer 12

Now the completing the square.

\(\large\color{black}{ \displaystyle {\rm \color{red}{a}}x^2+ {\rm \color{green}{b}}x+ {\rm \color{blue}{c}}=0 }\)


\(\large\color{black}{ \displaystyle {\rm \color{red}{a}}x^2+ {\rm \color{green}{b}}x= -{\rm \color{blue}{c}} }\)


\(\large\color{black}{ \displaystyle x^2+ \frac{{\rm \color{green}{b}}}{{\rm \color{red}{a}}}x= -\frac{{\rm \color{blue}{c}}}{{\rm \color{red}{a}}} }\)


\(\large\color{black}{ \displaystyle x^2+ \frac{{\rm \color{green}{b}}}{{\rm \color{red}{a}}}x+\left( \frac{{\rm \color{green}{b}}}{2{\rm \color{red}{a}}} \right)^2= -\frac{{\rm \color{blue}{c}}}{{\rm \color{red}{a}}}+\left( \frac{{\rm \color{green}{b}}}{2{\rm \color{red}{a}}} \right)^2 }\)


\(\large\color{black}{ \displaystyle \left( x+ \frac{{\rm \color{green}{b}}}{2{\rm \color{red}{a}}} \right)^2 = -\frac{{\rm \color{blue}{c}}}{{\rm \color{red}{a}}}+ \left( \frac{{\rm \color{green}{b}}}{2{\rm \color{red}{a}}} \right)^2 }\)


\(\large\color{black}{ \displaystyle \left( x+ \frac{{\rm \color{green}{b}}}{2{\rm \color{red}{a}}} \right) = \pm \sqrt{-\frac{{\rm \color{blue}{c}}}{{\rm \color{red}{a}}}+ \left( \frac{{\rm \color{green}{b}}}{2{\rm \color{red}{a}}} \right)^2} }\)


\(\large\color{black}{ \displaystyle x = -\frac{{\rm \color{green}{b}}}{2{\rm \color{red}{a}}}\pm \sqrt{-\frac{{\rm \color{blue}{c}}}{{\rm \color{red}{a}}}+ \left( \frac{{\rm \color{green}{b}}}{2{\rm \color{red}{a}}} \right)^2} }\)

Answer 13

I am not a lover of an abstract math (unless it is an optimization problem).

Answer 14

Well, you really did not need to go into detail about being , "Honorary Professor of Mathematics", but I just so assumed that you weren't poor in math. Therefore, it should be considered a compliment.

Thank you very much for your visual form of the formula. I guess this is more than a better way to view each of the equations that I am looking into. I now have more than one idea on how to view this equation.

Thank you @SolomonZelman. I will use your graphic and very detailed examples. :)

Answer 15

it is easier dealing with a specific example (with normal numbers, not a b c)
with a b c it is a rearrangement, with regular numbers it would be a solution.

Answer 16

you welcome though.