Finding Quadratics from Their Zeroes (page 1 of 2) Sections: Finding quadratics from their zeroes, Finding polynomials from a list of values Suppose I have some polynomial and I have to solve it. First I have to factor the polynomial, and then I have to solve each of the factors. For example:
First, I'll apply the Rational Roots Test to get a list of possible zeroes. Then I'll find some actual zeroes by testing the possibilities with synthetic division, and finally I'll end up with: (x + 3)(x – 2)(x + 1)(x – 1) = 0 Solving, I get a list of zeroes: x = ±1, –3, 2 I got these solutions by solving the factors. That's "working frontwards". I can also "work backwards" from the solutions. For instance, for x = 2 to be a solution, then I must have solved the factor equation x – 2 = 0, which means that x – 2 must have been a factor. The factors we find by working backwards from the zeroes are always of the form "(variable) minus (the given zero)". Having a factor of "(variable) minus (value)" means the same thing as having a solution of "(variable) equals (value)"; that is, if "x – a" is a factor, then "x = a" is a solution, and vice versa. We use this fact to find quadratics from their roots.
If the zeroes are at x = 4 and at x = –5, then, subtracting, the factor equations were x – 4 = 0 and x – (–5) = x + 5 = 0. Then the factors were x – 4 and x + 5 . Any factorable quadratic is going to have just the two factors, so these must be them. Then the original quadratic was something like: (x – 4)(x + 5) = x^{2} – 4x + 5x – 20 = x^{2} + x – 20
Why did I say that the quadratic
would be "something like" x^{2}
+ x – 20? Because
they may have divided something out when they solved the original quadratic.
For instance, if I had to solve 2x^{2}
– 2 = 0, I would first
divide off the 2
to get x^{2}
– 1 = (x + 1)(x – 1) = 0,
with solutions at x
= ±1. But if I multiply
back only the factors with variables, I get (x
– 1)(x + 1) = x^{2} – 1,
which is not quite what I started with. Any number of other quadratics
would also share those same two zeroes, including But that distinction, between "a" and "the", can be very important.
Subtracting, I get x – 1 = 0 and x – 3 = 0, so the factors were x – 1 and x – 3. However, I can't tell if they divided anything off to get the listed solutions. For instance, they may have started with factors like 4x – 4 or 5x – 15; I can't tell from only the zeroes. The general form of the family of quadratics (that is, the formula for every possible quadratic that has these zeroes) has to include any possible dividedoff numbers. For this, I will use a letter to represent the dividedoff numbers. The factors I know about are x – 1 and x – 3. Multiplying them together gives me x^{2} – 4x + 3. Since some number may have been divided out, I have to multiply it back in: The general form is a(x^{2} – 4x + 3) If they didn't divide anything out, then a will equal 1. By the way, your text may use some other letter for the constant; as long as you use some letter near the beginning of the alphabet (it's traditional), you should be fine.
Aha! With that extra point, I can narrow down the exact formula for the quadratic. The third point lets me account for that multiplier "a". I already know (from the previous exercise) that the general form of any quadratic with these zeroes is a(x^{2} – 4x + 3). But now I can plug in that point, and solve for a. Since (x, y) = (0, –6) is on the graph, then: a(x^{2}
– 4x + 3) = y
Then "the" quadratic is –2(x^{2} – 4x + 3) = –2x^{2} + 8x – 6
Since the zeroes are x = ^{3}/_{2} and x = ^{–5}/_{4}, then the factors were x – ^{3}/_{2} and x – ( ^{–5}/_{4} ) = x + ^{5}/_{4}. Or... maybe the factors were 2x – 3 and 4x + 5...? I really can't tell (not just from the zeroes, at least) whether they divided anything out when they were solving. So I'll need to use the point they gave me: a(x
– ^{3}/_{2} )(x + ^{5}/_{4}
) = y
Then the exact quadratic is y = –8(x – ^{3}/_{2} )(x + ^{5}/_{4} ) If your book or teacher doesn't like fractions, you can multiply through, like this: y = –8(x
– ^{3}/_{2} )(x + ^{5}/_{4} )
What if they ask you for a quadratic, but only give you one root?
When do I ever get square roots in my solutions to quadratics? When the stuff inside the square root in the Quadratic Formula doesn't simplify to a perfect square. And the Quadratic Formula always spits out those messy squareroot answers in oppositesigned pairs (from the "±" in front of the square root). So if x = sqrt(7) is a root, then the other root has to be x = –sqrt(7). This gives me my two factors: x – sqrt(7) and x – (–sqrt(7)) = x + sqrt(7). Multiplying, I get the quadratic x^{2} – 7. I tack on my "I don't know if you divided anything out" constant a to get the general solution a(x^{2} – 7). Plugging the point they gave me into this general form, I get: a(x^{2}
– 7) = y
So the quadratic is y = 3(x^{2} – 7) = 3x^{2} – 21 Top  1  2  Return to Index Next >>



Copyright © 20042012 Elizabeth Stapel  About  Terms of Use 




