Quadratics

A quadratic equation is an equation that is written in the form Ax^2+Bx+C, where A, B, and C are constants, and A does not equal 0. In many quadratic problems, the challenge is in finding x when the quadratic equals 0. How do we do this when the equation isn't linear? Well, there are a few methods...  

PART I: Factoring

Factoring is when you factor a quadratic expression into 2 linear expressions. For example, the quadratic equation 4x^2+21x+5 can be factored into (4x+1)(x+5).  A quadratic can be factored into these, and they can be solved separately easily, and thus, an intimidating second-degree expression is reduced into two pre-algebra problems.

Here, we see a factored form of a quadratic equation. Learning how to solve these is the first step in factoring quadratic equations. Think about this following problem.

According to your algebraic intuition, what x satisfies (x+5)(x-5)=0?

Solution: According to the zero rule, anything multiplied by 0 is 0. This is also the only way that two numbers can multiply to zero, if one of them is zero. If you are skeptical again, you can see for yourself by multiplying a few sets of numbers (you will see that anything multiplied by 0 is 0, and all things 0 have a factor of 0. Anyways, the key to this problem lies partly in realizing this property and partly in finding out what values of x make the two multiplied terms zero using algebra. To do this, all we have to do is solve a couple of linear equations. 

For x+5=0, subtracting 5 from both sides yields x=(-5)

For x-5=0, adding 5 to both sides yields x=5

We can substitute these values into the equation. For -5, (-5+5)(-5-5)=0*(-10)=0. For 5, (5+5)(5-5)=10*0=0. We can see that these values work. We can also expand the equation into x^2-25=0. Substituting again, we (-5)^2-25=25-25=0 and 5^2-25=25-25=0. -5 and 5 are the two values where (x+5)(x-5)=0. These two values are also called the roots of (x+5)(x-5), and in general, all values of x that make a single-variable polynomial 0 are called the polynomials roots. Therefore, if someone asked you what the roots of x^2-25 were, you'd reply "-5 and 5 are the roots".

Practice

1. For what values of x is (x+3)(x-2)=0?

2. What are the roots of (x+6)(x+2)?

3. What are the roots of (x-2)(x-5)?

4. Generally speaking, what are the roots of (x-a)(x-b)?

Enrichment

5. What value(s) of x satisfy x(x+5)=0?

6. What value(s) of x satisfy (x-10)^2=0?

7. What value(s) of x satisfy (x-8)(x+1)(2x-4)=0?

Challenge/Performance task

Write a polynomial in factored form with roots -5, -3, -2, 10, and 15.

Here are the answers:

1. Either x+3 is 0 or x-2 is 0. For the former one, x=0-3=-3 and for the latter one x=0+2=2. Therefore, the solutions to this equation are x=-3 and x=2.

2. Because the roots of an expression need to make it 0, either x+6 is 0 or x+2 is zero. Solving using linear equations, x is either equals -6 or -2.

3. Either x-2 or x-5 is 0. By using linear equations, we get x is either 2 or 5.

4. This problem is more abstract that the previous three, but it uses the same concept. x-a or x-b need to be 0 to make the multiplied product 0. We get x-a=0 and x-b=0 are the two solution. Adding a or b to both sides yields that for factored quadratic (x-a)(x-b)=0, the solutions are a and b. This rule will be crucial in the upcoming problems.

5.  This is an interesting problem because this is not directly in (x-a)(x-b) format for us. However, you can realize that x is simply (x-0), so the equation is converted into (x-0)(x-(-5))=0 where a=0 and b=(-5). Therefore, the values that satisfy this equation are 0 and -5.

6. In this problem, the quadratic is a square, unlike previous ones. However, y^2=y*y, so (x-10)^2=(x-10)(x-10), so a and b are both 10. Because of this, the only solution to this problem is 10.

7. This is a unique problem because there are three roots instead of two. Actually, this expression isn't even a quadratic, as all quadratics have two x factors, but that is irrelevant. We can extend our rule to the equation (x-a)(x-b)(x-c)=0. You can see that if x=c, then the equation can be simplified to