Completing the square
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Solving Quadratic Equations by Square Roots
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Example: Solving simple quadratic
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Solving quadratics by taking the square root
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Solving Quadratic Equations by Completing the Square
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Completing the square (old school)
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Example: Completing perfect square trinomials
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Example 1: Completing the square
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Example 2: Completing the square
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Example 3: Completing the square
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Example 4: Completing the square
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Example 5: Completing the square
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Completing the square 1
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Completing the square 2
Example 4: Completing the square Completing the Square 4
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- Complete the square on the general quadratic equation.
- We have ax squared plus bx plus c is equal to 0.
- So whenever I complete the square, actually whenever I
- deal with any of these types of quadratic equations, I
- always like to not have an a, or a non 1 coefficient, on the
- x squared terms. So let's make it into a 1 coefficient.
- And the easiest way to do that is just divide
- everything by a.
- So we divide every term on the left side by a, and of course
- we have to divide the right side by a as well.
- And so the left side will become x
- squared plus b over ax.
- And then I'll write c over a over here.
- So we have some room to add and subtract things so we can
- really complete the square.
- So plus c over a is equal to 0 divided by a, which is just
- going to be equal to 0.
- Now when we complete the square, we've seen this
- multiple times before, what we want to do is take the
- coefficient on the x term right here, it's b over a,
- take half of it.
- This right here is two times b over 2a.
- Right?
- The 2's cancel out.
- This is just 2 times b over 2a.
- So you take half of it, half of b over a is b over 2a.
- You take half of it, and then you square it, and you add it
- right here.
- So plus b squared-- Let me write it this way.
- b over 2a squared.
- And of course I can't just add something to one side of the
- equation, that would change the equation.
- I also either have to add it to the other side, or just
- subtract it from the same side that I'm adding it to.
- So I'll also subtract the b over 2a
- squared just like that.
- Now, the whole point of doing this is so these first three
- terms right here are a perfect square trinomial.
- That's what completing the square is all about.
- And we've seen the pattern multiple times.
- If I have, let's say, m plus n squared-- and I'm using m and
- n so we don't get confused with the a's and b's
- and x's over here.
- But if I have m plus n squared, we've seen multiple
- times, that's going to be equal to m squared plus 2mn
- plus n squared.
- And here we have that pattern now.
- That's the whole point behind completing the square.
- That's the whole point behind taking half of b over a--
- that's b over 2a-- and then squaring it and adding it
- right here.
- We now fit that pattern.
- m is x.
- n is b over 2a.
- And 2mn, if I take an x times a b over 2a, and multiply that
- by 2, I get b over ax.
- So this expression, right here, this trinomial, the
- first three terms, it is a perfect square trinomial, and
- we can write it as x plus b over 2a squared.
- And then of course we have all this other
- business right here.
- And then of course we have all of this
- other stuff right here.
- Which is negative b over, and let me just actually
- square it for you.
- So b over 2a squared is negative b
- squared over 4a squared.
- And then I have this plus c over a.
- But let's write it with the same denominator here.
- So I could have c over a, or I could multiply the numerator
- and the denominator by 4a.
- So if I multiply the numerator by 4a, I get 4ac.
- If I multiply the denominator by 4a, I get 4a squared.
- And the whole reason why I multiplied the numerator and
- the denominator by 4a was so that we have the same
- denominator right here.
- And, of course, that is going to be equal to 0.
- And we could simplify it a little bit more, or actually,
- well yeah we'll just simplify it next in the next step.
- We don't want to skip too many steps here.
- So you have x plus b over 2a squared.
- And then we could say, plus-- we could put the 4ac first, so
- we could say-- actually let's just say, plus negative b
- squared, plus 4ac, all of that over 4a squared is equal to 0.
- I didn't put the 4ac first, I just put the negative b
- squared there.
- Now let's isolate this squared binomial on
- the left hand side.
- And the easiest way we can do that is to subtract this thing
- from both sides of the equation.
- So let's do that.
- So you can imagine if we add b-- let me do this in a
- different color.
- If we were to add positive b squared minus 4ac over 4a
- squared on the left hand side, those will cancel out and
- we're also going to add it on the right hand side.
- Positive b squared minus 4ac over 4a squared.
- Anything I do the left I have to do the right.
- What do we end up with?
- We end up with, on the left hand side, these two guys
- cancel out.
- We have the same denomenator, when you add the numerators,
- that cancels with that.
- The 4ac cancels with the negative 4ac, these just
- completely cancel out.
- And on the left hand side, you just have x
- plus b over 2a squared.
- And on the right hand side, you have that being equal to b
- squared minus-- let me do that in a blue color.
- b squared minus 4ac, all of that over 4a squared.
- Now the next thing we probably want to do, if we want to
- really solve for x, is to take the square root of both sides
- of this equation.
- So let's do that.
- Let's take the square root of both sides of this equation.
- And if, when we do that-- We don't want to only take the
- positive square root because x plus b be over 2a could be a
- negative number, or it could be a positive number.
- So we want to take the positive and
- negative square root.
- So we could say that the square root, we could put the
- positive or negative here or, since we're taking the square
- root of both sides, we could put the
- positive or negative there.
- If you put the positive or negative on both sides, it's
- really just telling you the same thing.
- It really is all the different combinations.
- If the negative square root over here equals the negative
- square root over here, then it's just another combination
- of the different positives and negatives.
- So you could just write it as, this square root is equal to
- the plus or minus square root of b squared
- minus 4ac over 4a squared.
- Now what does this simplify to?
- Well, the left hand side just becomes x plus b over 2a is
- equal to-- and now it gets interesting.
- And you might even start recognizing parts of it.
- So let's take the plus or minus square root of the top.
- What is that going to be?
- And you could just take the plus or minus only of the top
- because, once again, the same principles apply.
- There's no reason why you have to do a plus or minus over a
- plus and minus and a plus or minus on the left hand side.
- There's only one combination here, where there's only one
- plus or minus on the numerator.
- I apologize if that confuses you.
- So let's write this as the plus or minus square root of b
- squared minus 4ac over-- What's the square
- root of 4a a squared?
- Well, it's just going to be 2a.
- Right?
- The square root of 4 is 2.
- The square root of a squared is a.
- And we're almost there.
- To solve for x, we just have to subtract b over 2a from
- both sides.
- We just have to subtract b over 2a from both sides of
- this equation.
- The left hand side, we just end up with our x.
- And then the right hand side, we have a negative b over 2a
- plus or minus the square root of b squared minus 4ac-- all
- of that over 2a.
- Since we have the same denominator, we can write this
- as negative b plus or minus the square root of b squared
- minus 4ac-- all of that over 2a.
- And we're done.
- We've solved for the x's and you see there's actually two
- solutions here.
- There's one where you take the positive square root, and
- there's another solution where you take the
- negative square root.
- If this square root exists, and if the positive and
- negative-- and if it's not 0, you're
- going to have two solutions.
- And this, right here, this result we have is-- Look, you
- give me any quadratic equation, you give me the a,
- the b, and the c, we could now substitute it into this
- formula essentially we just derived right here, and I'll
- give you the roots, I'll give you the x's for that quadratic
- equation-- the x's that satisfy
- that quadratic equation.
- And this formula, right here, for solving any quadratic
- equation is called the quadratic formula.
- And you could see it just comes straight out of
- completing the square.
- There's no mystery, magic here.
- But it's easily one of the most useful formulas in
- mathematics.
- And I'm usually not a huge proponent
- of memorizing things.
- But it probably will benefit you in life if you did.
- Hope you enjoyed that.
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