Finding Three Quadratic Polynomials That Each Have Two Zeros
Quadratic polynomials are among the most fundamental objects in algebra, and understanding how to shape them so that they possess exactly two distinct zeros is a key skill for students, teachers, and anyone working with polynomial functions. In this article we will walk through the theory behind zeros of quadratics, show a reliable method for constructing them, and then present three quadratic polynomials that have 2 zeros each. Each example will be accompanied by a short verification so you can see the zeros clearly.
Introduction
A quadratic polynomial has the general form
[ f(x)=ax^{2}+bx+c,\qquad a\neq0 . ]
Its zeros (also called roots or solutions) are the values of (x) for which (f(x)=0). Because of that, because a quadratic is a degree‑2 polynomial, the Fundamental Theorem of Algebra guarantees exactly two zeros in the complex number system—counting multiplicity. When we speak of “two zeros” in a typical high‑school context we usually mean two distinct real zeros, which occurs when the discriminant (b^{2}-4ac) is positive.
The goal of this section is to give you a clear, step‑by‑step recipe for building quadratics with two real zeros, and then to apply that recipe to produce three concrete examples.
Understanding Quadratic Polynomials and Their Zeros
Before we jump into construction, let’s review the key concepts that control the number and type of zeros.
| Concept | Formula / Condition | What It Tells You |
|---|---|---|
| Discriminant | (\Delta = b^{2}-4ac) | (\Delta>0) → two distinct real zeros; (\Delta=0) → one real zero (double root); (\Delta<0) → two complex conjugate zeros |
| Vertex form | (f(x)=a(x-h)^{2}+k) | The vertex is ((h,k)); the sign of (a) tells whether the parabola opens up ((a>0)) or down ((a<0)) |
| Factored form | (f(x)=a(x-r_{1})(x-r_{2})) | (r_{1}) and (r_{2}) are the zeros; expanding gives the standard form |
| Sum & product of roots | (r_{1}+r_{2}=-\frac{b}{a},; r_{1}r_{2}=\frac{c}{a}) | Useful for checking work or designing a quadratic from desired zeros |
If you already know two numbers you want as zeros, say (r_{1}) and (r_{2}), you can instantly write a quadratic in factored form and then expand it to standard form. This is the simplest way to guarantee exactly two zeros.
Step‑by‑Step Guide to Constructing Quadratics with Two Zeros
Follow these steps to create any quadratic polynomial that has two distinct real zeros.
-
Choose two distinct real numbers that you want as the zeros. Call them (r_{1}) and (r_{2}).
Example: (r_{1}=3), (r_{2}=-2) But it adds up.. -
Write the factored form using a non‑zero leading coefficient (a).
[ f(x)=a(x-r_{1})(x-r_{2}) ]
The value of (a) controls the vertical stretch/compression and the direction the parabola opens, but it does not change the zeros. For simplicity you may set (a=1) unless you have a specific reason to pick another value. -
Expand the product to obtain the standard form (ax^{2}+bx+c).
Multiply out ((x-r_{1})(x-r_{2})) and then distribute (a). -
Verify the discriminant (optional but recommended).
Compute (\Delta=b^{2}-4ac). You should find (\Delta>0), confirming two distinct real zeros. -
Check the zeros by solving (f(x)=0) (factoring or quadratic formula) to ensure they match your original choices.
That’s it! By picking any two different real numbers and any non‑zero (a), you automatically get a quadratic with exactly two zeros It's one of those things that adds up..
Three Example Polynomials
Below are three quadratics built using the procedure above. Each one has two distinct real zeros, and we show the work that leads to the final polynomial.
Example 1: Zeros at (-4) and (5)
- Chosen zeros: (r_{1}=-4), (r_{2}=5).
- Factored form with (a=1):
[ f_{1}(x)=(x+4)(x-5) ] - Expand:
[ f_{1}(x)=x^{2}-5x+4x-20 = x^{2}-x-20 ] - Discriminant: (\Delta = (-1)^{2}-4(1)(-20)=1+80=81>0).
- Zeros: solving (x^{2}-x-20=0) gives ((x-5)(x+4)=0) → (x=5) or (x=-4).
[ \boxed{f_{1}(x)=x^{2}-x-20} ]
Example 2: Zeros at (\frac{1}{2}) and (-3) (with a stretch factor)
- Chosen zeros: (r_{1}=\frac{1}{2}), (r_{2}=-3).
- Choose a leading coefficient (a=2) to avoid fractions in the final expression (optional).
Factored form:
[ f_{2}(x)=2\Bigl(x-\frac{1}{2}\Bigr)(x+3) ] - Expand stepwise:
[ \Bigl(x-\frac{1}{2}\Bigr)(x+3)=x^{2}+3x-\frac{1}{2}x-\frac{3}{2}=x^{2}+\frac{5}{2}x-\frac{3}{2} ]
Multiply by (2):
[ f_{2}(x)=2x^{2}+5x-3 ] - Discriminant: (\Delta = 5^{2}-4(2)(-3)=25+24=49>0).
- Zeros: using the quadratic formula,
[ x=\frac{-5\pm\sqrt{49}}{2\cdot2}=\frac{-5\pm
