@@markrobinson9956 It is still fairly easy in those circumstances. The trick is finding the right integer to multiply the equation by to get a new a that is a perfect square and an even b. I encourage people to learn that part of completing the square since completing the square is the easiest way to learn the sign change in a quadratic inequality and is a quick way to stress that an absolute value is the square root of a square.
I like the formula.. its just one thing to remeber and if you do these infrequently...and to be honest i have never ran into one even in electronics..a formual is way easier to remember over the long term.
The only time I've ever used the quadratic formula outside of a math class was in physics, to solve an exam problem. Reading the problem, it surprised me so much that I nearly blurted out "It's a ******* quadratic equation!"
That middle step is slightly over complicated, if you write it out fully it is simpler to under stand. x² - 12x + 5 = 0 (x-12/2)² - 36 +5 = 0 {Drop down the sign separating the first two terms into the bracket and then minus from the outside of the bracket half of the coefficient of the second term squared} (x-6)²-31=0 (x-6)² = 31 {Sqrt} x-6 = ±√31 x = 6 ± √31 (x-6)²
Why have I never learned this. This makes a lot of sense. This does require a little more thinking as you need to figure out how to complete the square so I can see how the quadratic formula is better for some people but for me this seems better.
For those not aware, take a general quadratic equation in the form of ax^2 + bx + c = 0. Solve for x by completing the square and you get... the quadratic equation. So even "I'll take the easy way and use the quadratic equation" is just "I am completing the square, just skipping some steps and jumping to the end". Usually I find completing the square is faster anyway, and I don't have to worry I remembered the equation incorrectly.
By Descartes' rule of signs, two sign swaps tell us that there are either two positive roots, or zero positive roots. And if there are zero positive roots, this implies that the two roots are complex conjugates with equal positive real parts.
@@dougr.2398 Sign swaps of the coefficients. You put the terms in descending order by power of x, and look at the signs on each coefficient. Coefficients of zero don't count as a sign swap, and continue the status-quo of the previous sign. For example: x^4 - 8*x^3 + 5*x^2 + 6*x + 1 = 0 This has 2 sign swaps, since it swapped from +1, to -8, then to +5. So there are either 2, or 0 positive real solutions. This one in particular, has 2 positive real solutions and 2 negative real solutions.
If I get a quadratic equation ax^2 + bx + c = 0 and it is not factorable, I would use CTS only when a=1 I can let go of b being even, but a not being one makes this a tedious task
Completing the square is elegant. The quadratic formula is/can be derived by completing the square. But teach it to even an A-grade high school kid and he or she might actually hate you for it. I no longer wanna teach kids because all they want is shortcuts and not understanding. I get complaints from students not understanding what I am teaching. I only talk math with those who love math and wanna understand math deeper. That’s my experience teaching and tutoring math both in Asia and in America. Anyone reading this, I wonder what your experience is-if you’ve been teaching math long enough?
you can look at the perfect square (x + a)^2 = x^2 + 2ax + a^2 that x-term is the 2ax term. the a is the thing we're looking for. we can see the coefficient of the x-term which will be the 2a, which here it is 12. then we divide it by 2 to get the a. Which then we add a^2 to both sides, to complete the square!
because we know that (a±b)^2 = a^2 ± 2ab + b^2(+2ab if a+b and -2ab if a-b) and the intention of this formula is to write the side with x in a form of perfect square with that being said, here, we have x^2 + 12x on the LHS Here, we want to write the LHS in a way so that it forms a perfect square expression x^2 + 12x = x^2 + 2*x*6 6^2 is the final piece of the puzzle to making the LHS a perfect square, so we add 6^2 on both sides to that the LHS can be written in a perfect square form
yes, but one requires you to memorise a formulae and the other is understanding the process to get the formulae. Understanding completing the square is important, but if you just need to solve the equation, the formulae is the easy end solution
@@aleycolt3824 People ask me why I use the Quadratic Formula, when I could just guess at possible roots and test them. I don’t bother with guessing. The Quadratic Formula 𝗮𝗹𝘄𝗮𝘆𝘀 works.
It's 12x. When you resolve (x+a)^2 you get x^2+2ax+a^2. The 2ax comes from the multiplication of x and a twice. Since your trying to reverse the bracket expansion, you need to look at the 2ax to find the value of a. Hope this helps
The idea is to start with the form: a*x^2 + b*x + c = 0 And turn it into the vertex form: a*(x - h)^2 + k = 0 Expand: a*x^2 - 2*a*h*x + a*h^2 + k = 0 Match coefficients: a = a b = -2*a*h c = a*h^2 + k Solving for h and k we get: h = -b/(2*a) k = c - a*h^2 = c - b^2/(4*a) Plug in a = 1 and b = -12, for this problem: h = +6 k = -31 Thus the vertex form solution is: (x + 6)^2 - 31 = 0
It is called "completing the square" because you are trying to make something of the form m^2 + 2mn + n^2, which equals (m + n)^2 and is called a perfect square. Dividing the b (12 in this case) by 2 and then adding the square of the quotient completes the perfect square.
@@Chbaxter_ For purposes of my comment, consider x^2 + 12*x + 5 = y. It's easier to explain with a positive b-term. Consider an actual square of side length x. Consider as well: as a rectangle of sides 12 by x, and 5 unit squares. We'd like to assemble this, so it is as close to a square as practical. Cut the 12 by x rectangle in half, so it becomes two 6 by x rectangles. Add it onto the x by x square, on perpendicular sides. Now we ALMOST have a square of side lengths (x + 6), but we're still missing part of it. We need 36 unit squares to complete it, but we only have 5. So add 31 unit squares to both sides, to get a perfect square on the left. (x + 6)^2 = y + 31 And when y=0: (x + 6)^2 = 31, which we can now directly solve for x.
A perfect square, (x - h)^2 expands as: x^2 - 2*h*x + h^2 We need to line up the given equation with the perfect square identity. In order for this to happen, -2*h needs to equal the given -12 coefficient. This means h = -12/2, which is -6.
I'll explain why, using x^2 + 12*x + 5 = 0 as an example. The idea is the same, but it's more intuitive with a plus sign in front of 12. Consider how each term appears as shapes. x^2 is a square of side length x 12*x is a rectangle, 12 by x. 5 is 5 individual 1 by 1 unit squares Cut the 12*x rectangle in half, so it is 2 rectangles that are each 6*x. Attach each of these rectangles to perpendicular sides of our x by x square. We now ALMOST have a square, with side length (x + 6). But something is missing. To complete this square, we need a 6*6 square, which is 36 unit squares. So add 36 to both sides x^2 + 12*x + 36 + 5 = 36 Now replace the perfect square: (x + 6)^2 + 5 = 36 And shuffle everything left: (x + 6)^2 - 31 = 0
Then you'll simply have a fraction with a denominator of 2, in your parenthetical expression. You'll also end up generating a fraction with a denominator of 4, unless you start with a special case that could cancel it out.
An application of quadratic formulas, I've used in real life is as follows. I've simplified the problem and changed the numbers, but the idea is the important part. A beam is 10 meters long, and will carry a uniform load, and will be supported at two points along its length. Determine the location of the two points, that will maximize the strength of the beam. After setting this up from Euler beam theory, and equating the criteria that matters for maximizing strength, you end up with: w*c^2/2 = w/2*((L - 2*c)^2/4 - c^2) where w is the uniform load per unit length c is the cantilever, and L is the total length w/2 cancels out, so we're left with: c^2 = (L - 2*c)^2/4 - c^2 Shuffle to one side, and expand: -c^2 - L*c + L^2/4 = 0 And, solving with the quadratic formula, you'll get: c = L/2*(sqrt(2) - 1) So for L = 10 meters, c = 2.07 meters
While this is true, I think Completing The Square is actually easier to understand -- whether it be through an area model or using the form x^2 + 2ax + a^2 + b or the related one, where x^2 is greater than one.
The point is to understand where the quadratic formula comes from, rather than to just see it as a magic equation without understanding it. There are also other applications of completing the square. For instance, integrating 1/(x^2 + 2*x + 5) dx If we had 1/(x^2 + 1), the solution would be arctan(x) + C. So we'll need to rig the integral, to make it look like this as closely as possible. Completing the square gives: 1/((x + 1)^2 + 1) dx Almost there. The integrand is just shifted horizontally from 1/(x^2 + 1), by 1 unit to the left. So the result will just be shifted by 1 unit to the left. Result: arctan(x + 1) + C
Because your goal is to make a perfect square, using x^2 and -12*x and an undetermined constant. I'll explain assuming the sign in front of the 12 is plus, but the idea is still the same for a negative. Start with an actual square of side length x. Now put two identical rectangles on two perpendicular sides of this square, to build a larger square. In order for this to happen, these rectangles will have side lengths 6*x, so they both add up to 12*x. Almost there. We now have an x^2 square, two 6*x rectangles, but we still have one more square to add. That square would be 36 unit squares, to fill a square that is 6 units by 6 units. This means: x^2 + 12*x + C can equal (x + 6)^2 when C = 36
easiest way to solve quadratic equation 👍
Cts is way faster if you know how to do it.
Unless a is not one or b is odd.
@@markrobinson9956 It is still fairly easy in those circumstances. The trick is finding the right integer to multiply the equation by to get a new a that is a perfect square and an even b. I encourage people to learn that part of completing the square since completing the square is the easiest way to learn the sign change in a quadratic inequality and is a quick way to stress that an absolute value is the square root of a square.
formula is easier
If the b term is an even number, i prefer CTS; if its uneven, Quadratic Formula to avoid fractions.
For anyone who does not get the joke, they are the same method:
ax^2 + bx + c = 0
ax^2 + bx = -c
x^2 + bx/a = -c/a
x^2 + bx/a + (b^2)/(4a^2) = -c/a + (b^2)/(4a^2)
[x + b/(2a)]^2 = (b^2 - 4ac)/(4a^2)
x + b/(2a) = ±[(b^2 - 4ac)^(1/2)]/(2a)
x = [-b ± (b^2 - 4ac)^(1/2)]/(2a)
Also, if the b is odd, multiply in a 4 then add b^2
x^2 - 13x + 5 = 0
x^2 - 13x = -5
4x^2 - 52x = -20
4x^2 - 52x + 169 = -20 + 169
(2x - 13)^2 = 149
2x - 13 = ±149^(1/2)
2x = 13 ± 149^(1/2)
x = [13 ± 149^(1/2)]/2
This is good also.
💀
Yes, CTS is a proof of the quadratic formula
Yes, CTS is a proof of the quadratic formula
I like the formula.. its just one thing to remeber and if you do these infrequently...and to be honest i have never ran into one even in electronics..a formual is way easier to remember over the long term.
The only time I've ever used the quadratic formula outside of a math class was in physics, to solve an exam problem. Reading the problem, it surprised me so much that I nearly blurted out "It's a ******* quadratic equation!"
They are one and the same anyways. The formula itself is derived from completing the square.
That middle step is slightly over complicated, if you write it out fully it is simpler to under stand.
x² - 12x + 5 = 0
(x-12/2)² - 36 +5 = 0 {Drop down the sign separating the first two terms into the bracket and then minus from the outside of the bracket half of the coefficient
of the second term squared}
(x-6)²-31=0
(x-6)² = 31
{Sqrt}
x-6 = ±√31
x = 6 ± √31
(x-6)²
I prefer to complete the square because I can retain the solution as a surd and avoid messy decimals. Also, it helps to work out the turning point.
Why have I never learned this. This makes a lot of sense. This does require a little more thinking as you need to figure out how to complete the square so I can see how the quadratic formula is better for some people but for me this seems better.
I learned completing the square once, my teacher hated it so we avoided it. Now it looks so useful.
If a=1 or a and b are both perfect squares, then I will use completing the square. If not, then i will use the quadratic formula.
On comparing x²-12x+5=0 with ax²+bx+c=0, we get a=1,b=-12,c=5
》x=(-b±(b²-4ac)½)/2a
x= ( -(-12)±((-12)²-4(1)(5))½ )/2(1)
x= (12±124½)/2
x=6±31½
That was amazing!!!!!!! Where you when I was going to high school?!?!?!? Great video, thanks!!!!!
That is best solved with the pq-formula.
Where do you believe the p-q formula is derived from? He basically derived it for you.
It is very refreshing to see someone speaking good English explaining math.
Why do you need someone speaking good English?
Good share, Trumpie, thanks.
ok trumpie
This man proves the genius of hard work.
For those not aware, take a general quadratic equation in the form of ax^2 + bx + c = 0. Solve for x by completing the square and you get... the quadratic equation. So even "I'll take the easy way and use the quadratic equation" is just "I am completing the square, just skipping some steps and jumping to the end". Usually I find completing the square is faster anyway, and I don't have to worry I remembered the equation incorrectly.
You could add that both roots are positive. (By inspection!)
By Descartes' rule of signs, two sign swaps tell us that there are either two positive roots, or zero positive roots. And if there are zero positive roots, this implies that the two roots are complex conjugates with equal positive real parts.
@@carultchsign swaps of what?
@@dougr.2398 Sign swaps of the coefficients. You put the terms in descending order by power of x, and look at the signs on each coefficient. Coefficients of zero don't count as a sign swap, and continue the status-quo of the previous sign.
For example:
x^4 - 8*x^3 + 5*x^2 + 6*x + 1 = 0
This has 2 sign swaps, since it swapped from +1, to -8, then to +5. So there are either 2, or 0 positive real solutions. This one in particular, has 2 positive real solutions and 2 negative real solutions.
Perfect
As an Indian , its very easy 😂
(Discriminant)
Bye the way thanks 🙏
Thank you dear teacher !
Very slowly method.
The true easiest way is the middle term factorisation
If I get a quadratic equation ax^2 + bx + c = 0 and it is not factorable, I would use CTS only when a=1
I can let go of b being even, but a not being one makes this a tedious task
Completing the square is elegant. The quadratic formula is/can be derived by completing the square. But teach it to even an A-grade high school kid and he or she might actually hate you for it. I no longer wanna teach kids because all they want is shortcuts and not understanding. I get complaints from students not understanding what I am teaching. I only talk math with those who love math and wanna understand math deeper. That’s my experience teaching and tutoring math both in Asia and in America. Anyone reading this, I wonder what your experience is-if you’ve been teaching math long enough?
Good to use formula
Aarigato Gozaimasu Sensei
What if he is Chinese...?! 😄
@@anujmarchande He's Korean. I think he knows what Aarigato Gozaimasu Sensei means, and still is honored to hear it.
x²-12x+5=0
x²-12x+36=31
(x-6)²=31
|x-6|=√31
x-6=±√31
x=6±√31 ❤❤
Ty
Quadratic formula🎉
Actually, I tend to prefer the Difference of Squares Method.
Quadratic equation is the easiest way to calculate it
But why it is not applicable on other type of quadratic equations 😢😢😢
Why 12 / 2 and then raised into power?
to make the equation on the right into a perfect trinomial
you mean on the left? @@saidaslamov8571
Please more question these type
Even 35 = 7x5 can give ans,
Quadratic formula
x^2-12x+5=0
x=[12+-rq(144-20)]/2=
[12+-rq124]/2=
[12+-2rq31]/2=
(6+-rq31)
I prefer completing the square cuz i always forget the formula
Dunno, I always go with "delta" formula.
I prefer the quadratic formula.
Why rule allows dividing the 12 by 2?
you can look at the perfect square
(x + a)^2 = x^2 + 2ax + a^2
that x-term is the 2ax term. the a is the thing we're looking for. we can see the coefficient of the x-term which will be the 2a, which here it is 12. then we divide it by 2 to get the a.
Which then we add a^2 to both sides, to complete the square!
If you using complex analysis method, you can appear be amazing
Why use two to divide the 12x? Is it because we’re simplifying the process or is there something I’m missing can someone explain plain lease
because we know that (a±b)^2 = a^2 ± 2ab + b^2(+2ab if a+b and -2ab if a-b) and the intention of this formula is to write the side with x in a form of perfect square
with that being said, here, we have x^2 + 12x on the LHS
Here, we want to write the LHS in a way so that it forms a perfect square expression
x^2 + 12x = x^2 + 2*x*6
6^2 is the final piece of the puzzle to making the LHS a perfect square, so we add 6^2 on both sides to that the LHS can be written in a perfect square form
Isn’t the Quadratic Formula the same as completing the square?
yes, but one requires you to memorise a formulae and the other is understanding the process to get the formulae.
Understanding completing the square is important, but if you just need to solve the equation, the formulae is the easy end solution
@@aleycolt3824 People ask me why I use the Quadratic Formula, when I could just guess at possible roots and test them.
I don’t bother with guessing. The Quadratic Formula 𝗮𝗹𝘄𝗮𝘆𝘀 works.
@@aleycolt3824formula
Mr. H in da house causing the good commotion
What house? Mad house??
What about The PQ
Why is 12 divided by 2?
the formula for CTS is (b/2)²
It's 12x.
When you resolve (x+a)^2 you get x^2+2ax+a^2. The 2ax comes from the multiplication of x and a twice.
Since your trying to reverse the bracket expansion, you need to look at the 2ax to find the value of a.
Hope this helps
I think so I still think that could have been slightly better explained for the slow among us. But thank you have a nice day
@@danieldare2640 the problem with most math teachers.
Completing The Square. I can't stand the QF.
how did we get (x-6)^2
The idea is to start with the form:
a*x^2 + b*x + c = 0
And turn it into the vertex form:
a*(x - h)^2 + k = 0
Expand:
a*x^2 - 2*a*h*x + a*h^2 + k = 0
Match coefficients:
a = a
b = -2*a*h
c = a*h^2 + k
Solving for h and k we get:
h = -b/(2*a)
k = c - a*h^2 = c - b^2/(4*a)
Plug in a = 1 and b = -12, for this problem:
h = +6
k = -31
Thus the vertex form solution is:
(x + 6)^2 - 31 = 0
i didnt know this existed
how comes over 6?
Why 12 Divided by 2??
I have the same question
It is called "completing the square" because you are trying to make something of the form m^2 + 2mn + n^2, which equals (m + n)^2 and is called a perfect square. Dividing the b (12 in this case) by 2 and then adding the square of the quotient completes the perfect square.
@@Chbaxter_ For purposes of my comment, consider x^2 + 12*x + 5 = y. It's easier to explain with a positive b-term.
Consider an actual square of side length x. Consider as well: as a rectangle of sides 12 by x, and 5 unit squares. We'd like to assemble this, so it is as close to a square as practical.
Cut the 12 by x rectangle in half, so it becomes two 6 by x rectangles. Add it onto the x by x square, on perpendicular sides.
Now we ALMOST have a square of side lengths (x + 6), but we're still missing part of it. We need 36 unit squares to complete it, but we only have 5. So add 31 unit squares to both sides, to get a perfect square on the left.
(x + 6)^2 = y + 31
And when y=0:
(x + 6)^2 = 31, which we can now directly solve for x.
Why do you divide 12 by 2?
A perfect square, (x - h)^2 expands as:
x^2 - 2*h*x + h^2
We need to line up the given equation with the perfect square identity. In order for this to happen, -2*h needs to equal the given -12 coefficient.
This means h = -12/2, which is -6.
Why he divide the 12 by 2?
I'll explain why, using x^2 + 12*x + 5 = 0 as an example. The idea is the same, but it's more intuitive with a plus sign in front of 12.
Consider how each term appears as shapes.
x^2 is a square of side length x
12*x is a rectangle, 12 by x.
5 is 5 individual 1 by 1 unit squares
Cut the 12*x rectangle in half, so it is 2 rectangles that are each 6*x.
Attach each of these rectangles to perpendicular sides of our x by x square. We now ALMOST have a square, with side length (x + 6). But something is missing.
To complete this square, we need a 6*6 square, which is 36 unit squares. So add 36 to both sides
x^2 + 12*x + 36 + 5 = 36
Now replace the perfect square:
(x + 6)^2 + 5 = 36
And shuffle everything left:
(x + 6)^2 - 31 = 0
Sir, why not to let the equal side, be zero?
because the eqatuion to the right is not a perfect square so impossible to get x= a number unless you use quadratic formula
@@saidaslamov8571 Thanks.
what if b is a odd number?
Then you'll simply have a fraction with a denominator of 2, in your parenthetical expression. You'll also end up generating a fraction with a denominator of 4, unless you start with a special case that could cancel it out.
Or you can use the quadratic formula
Lost me a quadratic. I golf for a living lol
The path of a golf ball is an application of quadratic formulas, because the shape of its flight is a parabola.
Me who is dumb asf in math back in the days and after i watched this i feel even dumber lmao. Can't understand shit.
The actual name of quadratic formula is Shri Dharacharyas formula...... He was an Indian mathematician who found this formula.
Everything isn't discovered by indians lmfao
nobodu care
@@dddaaa6965 I care about it bcz it's MADE IN INDIA
who cares
@@dddaaa6965 People who are proud abt their country care.... I think you are from a place where colonialism works
Where do you get the two to divide the 12? Why didn't it apply to anything else?
For what reason u need this in life?
An application of quadratic formulas, I've used in real life is as follows. I've simplified the problem and changed the numbers, but the idea is the important part.
A beam is 10 meters long, and will carry a uniform load, and will be supported at two points along its length. Determine the location of the two points, that will maximize the strength of the beam.
After setting this up from Euler beam theory, and equating the criteria that matters for maximizing strength, you end up with:
w*c^2/2 = w/2*((L - 2*c)^2/4 - c^2)
where w is the uniform load per unit length
c is the cantilever, and L is the total length
w/2 cancels out, so we're left with:
c^2 = (L - 2*c)^2/4 - c^2
Shuffle to one side, and expand:
-c^2 - L*c + L^2/4 = 0
And, solving with the quadratic formula, you'll get:
c = L/2*(sqrt(2) - 1)
So for L = 10 meters, c = 2.07 meters
The quadratic formula is derived by completing the square. They aren't different methodologies.
While this is true, I think Completing The Square is actually easier to understand -- whether it be through an area model or using the form x^2 + 2ax + a^2 + b or the related one, where x^2 is greater than one.
Completing the square wastes time
The point is to understand where the quadratic formula comes from, rather than to just see it as a magic equation without understanding it.
There are also other applications of completing the square. For instance, integrating 1/(x^2 + 2*x + 5) dx
If we had 1/(x^2 + 1), the solution would be arctan(x) + C. So we'll need to rig the integral, to make it look like this as closely as possible.
Completing the square gives:
1/((x + 1)^2 + 1) dx
Almost there. The integrand is just shifted horizontally from 1/(x^2 + 1), by 1 unit to the left. So the result will just be shifted by 1 unit to the left.
Result: arctan(x + 1) + C
Either
Why do you divide the 12 by 2?
Because your goal is to make a perfect square, using x^2 and -12*x and an undetermined constant.
I'll explain assuming the sign in front of the 12 is plus, but the idea is still the same for a negative.
Start with an actual square of side length x.
Now put two identical rectangles on two perpendicular sides of this square, to build a larger square.
In order for this to happen, these rectangles will have side lengths 6*x, so they both add up to 12*x.
Almost there. We now have an x^2 square, two 6*x rectangles, but we still have one more square to add. That square would be 36 unit squares, to fill a square that is 6 units by 6 units.
This means:
x^2 + 12*x + C can equal
(x + 6)^2
when C = 36
@@carultch Great explanation, thanks. I didn't get the idea of a perfect square, being an actual square lol. I thought it was just maths terminology 😇