How to Complete The Square - Step-by-Step Example….MUST Know to Solve Quadratic Equations 

Glad you emphasized getting the constant term on the right-hand side. I've seen sometimes where the method is to add and subtract the constant term on the same side, and it's confusing and an invitation to making mistakes.

Really great. Great job John, walking us thru CTS.

All one really has to do is (1 make the coefficient on the squared term unity, (2 get the constant term on the right-hand side, and (3 obtain 1/2 the coefficient on the "x" term and square it (also adding said squared term to the right-hand side).

So many steps needed to complete the square that it is easier to use the so-called abc-formula (including the Discriminant).

I have to admit that I have just defaulted to the quadratic formula since 8th grade when I learned it. Outside of school tests, I have never had to actually use the complete the square procedure although I am now having to teach it to my son.

Thanks. Even though I have a BS in Applied Mathematics this is new to me. I learned Viete’s Relations and the quadratic equation for solving any quadratic equation.

y= a(x - h)^2 - k h= .5(-b * a^-1) (a^-1 means reciprocal of the a value in this case it is also one 1/2) k= -a(h^2) - 1 finally now that we have the equation in vertex form we solve for x with: x= h + sqroot(-k/a) and x= h - sqroot(-k/a) x= -5/4 + (sqroot(33) / 4) and x= -5/4 - (sqroot(33) / 4 (man I wish we could type like how you can write out on paper. Sorry for the pseudo code lol) Basically what I did above is called Vertex form. It Can be found in other RU-vid videos. I use -1 as an exponent since you are actually supposed to divide by the a constant. And in case you didn't know dividing by a number even if it is a fraction, is the same as multiplying the first number by the reciprocal of the second number. (as and example 4/2 is the same as saying 4 * 1/2)

im going to have to go over this a few times. Wouldvit be correct to say that different forms of math have their own particular languages and formats?

Just find the value of discriminant if its worth finding solutions using perfect square method.

Since the QF is a direct computation from completing the square, why would we ever need completing the square over the QF? Using ax^2 + bx + c = 0 and completing the square gives the QF in the final answer,

I had a good teacher in high school 97% passed.

I don't understand why you would solve this equation by completing the square while there is much easier route: the modified ABC-formula in 4 steps... with in this case a = 2 b = 5 and c = -1 1. Determining the determinator: D = b² - 4ac = 25 - 4.(2).(-1) = 25 + 8 = +33 Positive number so there are two intersections of the parabola with the x-axis If D = 0 then there is just one intersection at xtop If D < 0 then there are no intersections to calculate. 2. The x-coordinate of the parabola can be found at xtop = - b / 2a = - (5) / 2.(2) = - 5 / 4 = - 1.25 The y-coordinate of the parabola can be found with ytop = f(xtop) = 2.( - 5/4 )² + 5.( - 5/4 ) - 1 = 50/16 - 100/16 - 16/16 = - 66/16 = - 33/8 = - 4 1/8 3. The distance of the intersections of the parabola with the x-axis to xtop is delta = V(D) / 2a = V33 / 2.(2) = V33/4 ~ 1.44 4. And the result is x = xtop -/+ delta = -5/4 -/+ V33/4 So x1 = ( - 5 - V33 ) / 4 ~ - 2.69 and x2 = ( - 5 + V33 )/ 4 ~ + 0.19 With these 4 steps you get 3 of the most important points of the parabola: ( - 2.69 , 0 ) and (- 1.25 , - 4.13 ) and ( 1.44 , 0 )

X=(-5-root 33)/4 X=(-5+root33)/4

At 10:34, The +/- is actually generated from the *LEFT* side of the equation, not the right side!!! The Sqrt( (x+5/4)^2 ) generates +/- ( x+5/4). The principal square root of (33/16) does *NOT* generate the +/- . Then to reduce the left side ... +/-(x+5/4) the +/- is first transferred, from the left side, to the right side. Then the parentheses, on the left side can be removed. Your explanation violated this rule ... The principal square root of a positive number is *always* a positive number !!!

What happened to the term 5/2x?

This problem justifies wordiness.