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Published byVincent Shepherd Modified over 4 years ago

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A quadratic equation is a second degree polynomial, usually written in general form: The a, b, and c terms are called the coefficients of the equation, and the x guy is the variable. Technically, the a coefficient is called the quadratic coefficient, the b is the linear coefficient, and the c is the constant term. For whatever reason, these equations are called quadratics, even though they’re always a second degree polynomial (“quad” means “four”).

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A solution to a quadratic equation is an x value such that the following equation is true: Such an x is called a root of the polynomial. Fortunately, there’s a formula, cleverly referred to as the quadratic formula, that allows us to find these roots. Here it is:

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The most difficult part about solving quadratic equations is remembering the quadratic formula; from there, it’s just a matter of plugging in numbers into it. Here it is again! (Even though it doesn’t explicitly say “1 x 2 + …” in the equation, the 1 is understood; it’s just like in grammar, when commands such as “get out” have an understood “you” in front: “(You) get out”). Let’s plug in our a, b, and c values:

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Time to simplify! If we choose + in our ± part: This will give us two answers, since the ± symbol means we have a choice between + and –. We can check that these are indeed the roots by plugging them back into our equation:

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In that example, the quadratic coefficient a was 1. Quadratic equations like that are called monic polynomials. Whether a quadratic equation is monic or not isn’t really a big deal, though it can come in handy if all the coefficients are divisible by the quadratic coefficient. What is really neat, though, is that any quadratic equation can be written as a monic one! All we have to do is divide everything by a : a This is really only needed to make the quadratic formula easier to solve, since in this case we have one less variable to plug in. Sometimes, it makes it harder, though, so you have to be careful!

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The discriminant of the quadratic formula tells us a lot about the number of roots the quadratic will have. The discriminant is this: If our discriminant is positive: We will have two distinct roots: If our discriminant is zero: We will have one root: If our discriminant is negative: We won’t have any real roots! Side note: if the roots are perfect squares (meaning the discriminant is of the form a 2 ), then our roots are rational numbers (fractions or whole numbers).

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When we know the roots of a quadratic equation, we are able to factor it. If the roots to a quadratic are u and v, then the quadratic can be factored into: ( x – u )( x – v ) This is something that “just is”. It makes sense mathematically, though it’s one of those things that are never derived. If we were to plug in u into the above expression, then we would end up with: ( u – u )( u – v ) = 0 ( u – v ) = 0 The same thing would happen if we plug in v : ( v – u )( v – v ) = ( v – u ) 0 = 0 This is fantastic, since we know a number is a root of a quadratic when plugging it in gives us zero. Yay!

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Factor the quadratic equation: 2 x 2 – 4 x – 6 = 0 In order to factor this guy, we need to find out what its roots are. That’s not bad; we just have to plug all the numbers into the quadratic formula and we’re done! But, all the coefficients of the equation are divisible by 2. In this case, it’s much easier to divide both sides of the equation by 2, and then plug in what’s left into the quadratic equation. 2 x 2 – 4 x – 6 = 0 2 Great! Now that we have this quadratic as a monic polynomial, we can solve for the x values that make this equation true.

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So, our roots are x = 3 and x = – 1. To factor this equation out, we just need to plug these numbers in as our u and v values from a couple slides ago, and then we’re home free. ( x – u )( x – v ) We can just put in 3 for u and – 1 for v : ( x – 3)( x – (-1)) = ( x – 3)( x + 1) That’s it! Factored out, 2 x 2 – 4 x – 6 = 2( x – 3)( x + 1). Time to plug the numbers into our quadratic formula: Now our a is 1, b = -2, and c = -3: Don’t forget the 2 we factored out before!

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Completing the square is probably the most difficult thing to do involving quadratic equations. After you complete the square, you will have something that will look like this: k ( x – a ) 2 + d Where k and d are constants. Notice what happens when we multiply ( x – a ) 2 out: ( x – a ) 2 = ( x – a ) ( x – a ) = x 2 – 2 ax + a 2 So, it appears that the middle b term is twice the number that appears in the completed –squares part. Also, our c term is just that number squared.

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When we’re out to complete the square, we have to look at the middle b term. That number will tell us everything we need to know about what the last number will be in order to complete the square. Example: Say we want to complete the square for x 2 + 4 x + 3. Our middle coefficient is 4. So, the number we’re looking for must be half that, so we want a 2: ( x – 2) 2 But, the last term in our equation is 3. We know it’s supposed to be 2 squared, which is 4. What do we do about this? There’s a nifty trick people like to use, where essentially we’re adding 0 to our expression (which of course is a perfectly legal move; adding zero to anything doesn’t change the expression at all). But, instead of just adding 0, which won’t help much, we can rewrite 0 as 1 – 1.

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So, our quadratic becomes x 2 + 4 x + 3 = x 2 + 4 x + 3 + (1 – 1) = ( x 2 + 4 x + 3 + 1 )– 1 = ( x 2 + 4 x + 4)– 1 = ( x + 2) 2 – 1 We’re done! That’s how you complete the square for any quadratic equation you like. Why 1 – 1? I’ve chosen 1 because we need 4, but we’re only given a 3. What do we do to make 3 equal to 4? We add 1 to it! But, we have to subtract 1 also, otherwise we’d be changing the expression (which is an extremely bad thing to do).

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