How to Build Quadratics from Roots?

A step-by-step guide to building quadratics from roots

If \(α\) and \(β\) are the two roots of a quadratic equation, the formula for constructing the quadratic equation is:

\(\color{blue}{x^2-(α + β)x + αβ = 0}\)

That is,

\(\color{blue}{x^2\:-\:\left(sum\:of\:roots\right)x\:+\:product\:of\:roots\:=\:0}\)

If a quadratic equation is given in the standard form, we can find the sum and product of the roots using the coefficients of \(x^2, x\), and the constant term.

Let’s consider the standard form of a quadratic equation,

\(ax^2+bx+c=0\)

Where \(a, b\) and \(c\) are real and rational numbers.

Let \(α\) and \(β\) be the two zeros of the quadratic equation above. Then the formula for obtaining the sum and the product of the roots of a quadratic equation is:

\(\color{blue}{α+β=-\frac{b}{a}=-\frac{coefficient\:of\:x}{coefficient\:of\:x^2\:}}\)

\(\color{blue}{αβ=\frac{c}{a}=\frac{constant\:term}{coefficient\:of\:x^2\:}}\)

Note: The irrational roots of a quadratic equation occur as conjugate pairs. That is if \((m\:+\sqrt{n})\) is a root, then \((m\:-\sqrt{n})\) is the other root of the same quadratic equation.

Build Quadratics from Roots – Example 1:

Build the quadratic equation whose roots are \(2\) and \(-\frac{1}{2}\).

Solution:

The sum of the roots is: \(2+ (-\frac{1}{2})= \frac {3}{2}\)

The product of the roots is: \( 2\times (-\frac{1}{2})= -1\)

Formation of the quadratic equation: \(x^2\:-\:\left(sum\:of\:roots\right)x\:+\:product\:of\:roots\:=\:0\)

So, the quadratic equation is: \(x^2- \frac{3}{2}x -1=0\)

Complex Roots and Quadratic Formulas

When roots are not nice integers or simple fractions, you’ll use the quadratic formula. Understanding the relationship between the formula and the factored form deepens your algebra skills.

Using the Quadratic Formula to Find Roots

For \(ax^2 + bx + c = 0\), the roots are given by:

\(x = \frac{-b ± \sqrt{b^2 – 4ac}}{2a}\)

Once you find the roots \(r_1\) and \(r_2\), you can immediately write the factored form:

\(f(x) = a(x – r_1)(x – r_2)\)

Example: Solve \(2x^2 – 5x – 3 = 0\) using the quadratic formula.

Here a = 2, b = -5, c = -3.

\(x = \frac{5 ± \sqrt{25 + 24}}{4} = \frac{5 ± \sqrt{49}}{4} = \frac{5 ± 7}{4}\)

So \(x = \frac{12}{4} = 3\) or \(x = \frac{-2}{4} = -\frac{1}{2}\).

Factored form: \(f(x) = 2(x – 3)(x + \frac{1}{2})\).

To verify, expand: \(2(x – 3)(x + \frac{1}{2}) = 2(x^2 + \frac{1}{2}x – 3x – \frac{3}{2}) = 2(x^2 – \frac{5}{2}x – \frac{3}{2}) = 2x^2 – 5x – 3\). ✓

When the Discriminant is Negative: Complex Roots

The discriminant is \(b^2 – 4ac\). If it’s negative, the square root involves imaginary numbers, and the roots are complex (non-real).

Example: Solve \(x^2 + x + 1 = 0\).

Discriminant: \(1 – 4 = -3\). The roots are:

\(x = \frac{-1 ± \sqrt{-3}}{2} = \frac{-1 ± i\sqrt{3}}{2}\)

These are complex roots. The quadratic \(x^2 + x + 1\) doesn’t cross the x-axis on a real graph, but in the complex plane, it still has roots.

Factored form in complex numbers: \(f(x) = (x – \frac{-1 + i\sqrt{3}}{2})(x – \frac{-1 – i\sqrt{3}}{2})\).

Quadratic Applications in Real Problems

Beyond algebra, quadratics model many real phenomena. Understanding how to build them from roots helps you set up these models.

Business: Profit and Revenue

A company’s profit as a function of units produced might be \(P(x) = -2x^2 + 100x – 500\). The break-even points (where profit = 0) are found by solving \(-2x^2 + 100x – 500 = 0\), or \(x^2 – 50x + 250 = 0\).

Using the quadratic formula: \(x = \frac{50 ± \sqrt{2500 – 1000}}{2} = \frac{50 ± \sqrt{1500}}{2} = \frac{50 ± 10\sqrt{15}}{2} = 25 ± 5\sqrt{15}\).

So the break-even points are approximately \(x \approx 5.6\) units and \(x \approx 44.4\) units. Below 5.6 or above 44.4 units, the company loses money.

Physics: Projectile Motion

A ball thrown upward has height \(h(t) = h_0 + v_0 t – \frac{1}{2}gt^2\), where \(h_0\) is initial height, \(v_0\) is initial velocity, and g is gravitational acceleration (about 9.8 m/s²). When does it hit the ground? Set h(t) = 0 and solve.

If \(h_0 = 2\) meters (thrown from a 2-meter platform) and \(v_0 = 20\) m/s, then:

\(0 = 2 + 20t – 4.9t^2\)

\(4.9t^2 – 20t – 2 = 0\)

Using the quadratic formula: \(t = \frac{20 ± \sqrt{400 + 39.2}}{9.8} = \frac{20 ± \sqrt{439.2}}{9.8} \approx \frac{20 ± 20.96}{9.8}\).

So \(t \approx 4.18\) seconds (positive, physically meaningful) or \(t \approx -0.098\) seconds (negative, not physical). The ball hits the ground after about 4.18 seconds.

Why Understanding Roots and Quadratics Matters

Quadratics are the simplest non-linear functions. They appear everywhere in applied mathematics. By mastering the ability to build them from roots, find their roots, and interpret them geometrically, you gain powerful problem-solving tools. You see that algebra isn’t just symbol manipulation—it’s a language for describing relationships between quantities. The roots of a quadratic are the solutions to real problems: when does profit break even? When does a projectile land? When do two objects meet?

Building quadratics from roots is building a bridge between abstract algebra and concrete reality. That bridge is where true mathematical understanding lives.