To find a quadratic expression when all you have are the zeroes (also called solutions or roots), you work backwards from the zeroes. Take a solution, x = a, and convert it into a factor, x − a.
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In other words, to get a quadratic from its zeroes, follow these steps:
To understand why this works, let's think back to how we have worked things in the other direction; that is, when we were finding the zeroes from the quadratic.
Suppose I have a quadratic and I have to solve it. First I have to factor the quadratic, and then I have to solve both of the factors. For example, suppose they ask me to find the solutions of the following quadratic:
x2 − 2x − 15
To find the roots of the quadratic, I'll factor it:
(x − 3)(x − 5)
Setting each factor equal to zero and solving, I get the two solutions:
x = −3, −5
I got these solutions by solving the quadratic's factors. That's working frontwards. To get the quadratic from the zeroes, I just work the above process in reverse; that is, I work backwards. Finding the zeroes from the quadratic and finding the quadratic from the zeroes are just two sides of the same coin.
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 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)
= x2 − 4x + 5x − 20
= x2 + x − 20
Why did I say that the quadratic would be "something like" x2 + x − 20? Because they may have divided out some common (numerical) factor when they solved the original quadratic.
For instance, if I had to solve 2x2 − 2 = 0, I would first divide off the 2 to get x2 − 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) = x2 − 1, which is not quite what I started with. Any number of other quadratics would also share those same two zeroes, including 5x2 − 5, 17x2 − 17, and 4 − 4x2.
If they only give me the zeroes, I can't tell if they divided anything out. The quadratic answer I gave in the problem above is good enough, though, because they only asked for "a" quadratic with the given zeroes, not "the" quadratic.
But that distinction, between "a" and "the", can be very important.
I convert the zeroes into factor form, and 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 divided-off numbers. For this, I will use a letter to represent the divided-off numbers.
The factors I know about are x − 1 and x − 3. Multiplying them together gives me x2 − 4x + 3. Since some number may have been divided out, I have to multiply it back in. So the general form is:
a(x2 − 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 (because it's traditional, for some reason), you should be fine.
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To find a quadratic from its zeroes and some other point (that is, some point off the x-axis), you convert the zeroes to factors, to get x2 + bx + c. Then you plug that extra point into the general quadratic equation, y = a(x2 + bx + c), where a is the unknown numerical multiplier from whatever was the original quadratic. Then solve for the value of a.
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(x2 − 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(x2 − 4x + 3) = y
a((0)2 − 4(0) + 3) = −6
3a = −6
a = −2
This tells me that the authors divided out a −2 before they found the zeroes; I'll need to multiply that −2 back onto the quadratic. Then "the" quadratic that they're looking for is:
−2(x2 − 4x + 3)
= −2x2 + 8x − 6
Since the zeroes are and , then the factors were and . 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 to find the exact actual original quadratic:
Okay; now I know the value of the constant that they'd divided out when they were solving the quadratic. With this information, I can multiply everything back together to get the original quadratic:
Then my answer is:
−8x2 + 2x + 15
While the factored form of the above quadratic (namely, ) is equal to the multiplied-out form, your instructor will almost certainly want you to give your answer in the multiplied-out form. If in doubt, provide both forms.
When you're asked to find a quadratic and you're given only one of the roots, this means that they're expecting you to use what you've learned about how the Quadratic Formula spits out roots in pairs, because of the "plus/minus" in front of the square root. In other words, when they've given you only the one zero (and probably another point not on an axis), that zero will be in a square root, and the other zero will be the same number but the opposite sign.
When do I ever get square roots in my solutions to quadratics? That only happens 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 square-root answers in opposite-signed pairs — from the "±" in front of the square root.
So if is a root, then the other root has to be . This gives me my two factors: and .
Multiplying these two factors together, I get the quadratic x2 − 7. I tack on my "I don't know if you divided anything out" constant a to get the general solution a(x2 − 7). Plugging the point they gave me into this general form, I get:
a(x2 − 7) = y
a((2)2 − 7) = (−9)
a(4 − 7) = −9
−3a = −9
a = 3
I'll multiply 3 through x2 − 7 to get the desired quadratic, which is:
3x2 − 21
You can use the Mathway widget below to practice finding the quadratic equation from its solutions. Try the entered exercise (being the two solution x-values), or type in your own exercise. Then click the button and select "Find the Quadratic Equation Given the Roots" to compare your answer to Mathway's. (Or skip the widget and continue on to finding polynomials from their zeroes.)
(Click "Tap to view steps" to be taken directly to the Mathway site for a paid upgrade.)