Quadratic Equations

A quadratic equation is one that can be written in the following form, called the canonical form:

\[ a x ^ 2 + b x + c = 0 \quad , \quad a \neq 0 \]

The real numbers \( a , b \) and \( c \) are called the quadratic coefficient, linear coefficient, and constant term.

It can always be assumed that the quadratic coefficient is greater than zero. In fact, if \( a < 0 \), we can simply multiply both sides by \( -1 \) to bring it to the case \( a > 0 \).

Table of Contents

• Completing the Square
• Reduced Formula
• Monomial Quadratic Equations
• Pure Quadratic Equations
• Spurious Quadratic Equations
• Relation Between the Sum and Product of Roots
• Solved Exercises
• Geometric Meaning

In this section, we will derive the general formula to solve any quadratic equation. We start from the canonical form:

\[ ax^2 + bx + c = 0, \quad a \neq 0 \]

To simplify the calculations, we divide everything by \( a \) so that the coefficient of the quadratic term becomes 1:

\[ x^2 + \frac{b}{a}x + \frac{c}{a} = 0 \]

Now, we isolate the constant term by moving it to the right-hand side:

\[ x^2 + \frac{b}{a}x = -\frac{c}{a} \]

At this point, we apply the completing the square method. The trick is to add and subtract the right term to turn the left side into a perfect square. This term is:

\[ \left(\frac{b}{2a}\right)^2 \]

We add it to both sides:

\[ x^2 + \frac{b}{a}x + \left(\frac{b}{2a}\right)^2 = -\frac{c}{a} + \left(\frac{b}{2a}\right)^2 \]

The left side is now the square of a binomial:

\[ \left(x + \frac{b}{2a}\right)^2 = \frac{b^2}{4a^2} - \frac{4ac}{4a^2} \]

We rewrite the right side with a common denominator:

\[ \left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a^2} \]

Now, we take the square root of both sides, remembering that the square root of a square is the absolute value:

\[ \left| x + \frac{b}{2a} \right| = \frac{\sqrt{b^2 - 4ac}}{2a} \]

From here, we directly solve for \( x \):

\[ x + \frac{b}{2a} = \pm \frac{\sqrt{b^2 - 4ac}}{2a} \]

Finally, we isolate \( x \) and obtain the famous quadratic formula:

\[ x_{1,2} = -\frac{b}{2a} \pm \frac{\sqrt{b^2 - 4ac}}{2a} \]

The term under the square root, known as the discriminant and denoted by \( \Delta \), is defined as:

\[ \Delta = b^2 - 4ac \]

But what does the discriminant represent? It allows us to quickly understand the type of solutions the equation will have. Let’s analyze it in the three possible cases:

• \( \Delta > 0 \): The discriminant is positive, so the square root is a real number. This means that the equation has two distinct real solutions.
• \( \Delta = 0 \): The square root of zero is zero, so the formula gives us a single repeated solution. In other words, the equation has two coincident solutions (or one double solution).
• \( \Delta < 0 \): The square root of a negative number is not a real number, so the equation has no real solutions, but two complex solutions with an imaginary part.

This means that by simply looking at the value of \( \Delta \), we can predict the nature of the solutions without having to solve the equation directly.

The reduced formula is a simplified version of the quadratic equation solution formula, useful when the coefficient \( b \) is even.

Consider a quadratic equation in its canonical form:

\[ ax^2 + bx + c = 0 \]

If the coefficient \( b \) is even, we can write it as:

\[ b = 2k \]

Substituting into the equation gives us:

\[ ax^2 + 2kx + c = 0 \]

The classic solution formula is:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Substituting \( b = 2k \):

\[ x = \frac{-2k \pm \sqrt{(2k)^2 - 4ac}}{2a} \]

\[ x = \frac{-2k \pm \sqrt{4k^2 - 4ac}}{2a} \]

\[ x = \frac{-2k \pm 2\sqrt{k^2 - ac}}{2a} \]

Dividing both the numerator and denominator by 2:

\[ x = \frac{-k \pm \sqrt{k^2 - ac}}{a} \]

Finally, we can express the reduced formula as:

\[ x = \frac{-\frac{b}{2} \pm \sqrt{\left(\frac{b}{2}\right)^2 - ac}}{a} \]

The reduced discriminant is given by:

\[ \Delta' = \left(\frac{b}{2}\right)^2 - ac \]

Now let's compare it with the discriminant of the full formula:

\[ \Delta = b^2 - 4ac \]

Substituting \( b = 2k \), we get:

\[ \Delta = (2k)^2 - 4ac \]

\[ \Delta = 4k^2 - 4ac \]

Dividing everything by 4:

\[ \frac{\Delta}{4} = k^2 - ac \]

Since \( k = \frac{b}{2} \), we can rewrite:

\[ \frac{\Delta}{4} = \left(\frac{b}{2}\right)^2 - ac \]

This is exactly the definition of \( \Delta' \).

Therefore, we can conclude that:

\[ \Delta' = \frac{\Delta}{4} \]

An equation is called monomial if it reduces to a single quadratic term, i.e., in the form:

\[ ax^2 = 0 \]

To solve this equation, we divide both sides by \( a \) (assuming \( a \neq 0 \)):

\[ x^2 = 0 \]

Taking the square root, we obtain the solution:

\[ x = 0 \]

Although the value is unique, mathematically, we consider two coincident solutions: \( x_1 = x_2 = 0 \).

An equation is called pure if, in the general form \( ax^2 + bx + c = 0 \), the coefficient \( b \) is zero, reducing to:

\[ ax^2 + c = 0 \]

To solve this equation, we move the constant term \( c \) to the other side:

\[ ax^2 = -c \]

We divide both sides by \( a \):

\[ x^2 = -\frac{c}{a} \]

Solutions exist only if \( \displaystyle -\frac{c}{a} \geq 0 \), otherwise the equation has no real solutions. If the value under the square root is positive, we obtain:

\[ x_{1,2} = \pm \sqrt{-\frac{c}{a}} \]

An equation is called spurious if the constant term is zero, i.e.:

\[ ax^2 + bx = 0 \]

In this case, we can solve it by factoring out \( x \) as a common factor:

\[ x (ax + b) = 0 \]

Applying the zero product property, we obtain the two solutions:

\[ x = 0 \quad \text{or} \quad x = -\frac{b}{a} \]

These solutions can also be found by applying the general solution formula for quadratic equations.

Consider the quadratic equation of the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are the coefficients. Let \( x_1 \) and \( x_2 \) be the roots of this equation. Now, we want to express the equation in terms of the roots. A quadratic equation can be written as the product of the factors \( (x - x_1) \) and \( (x - x_2) \), so we can write:

\[ a(x - x_1)(x - x_2) = 0 \]

Expanding the product on the left, we obtain:

\[ a(x^2 - (x_1 + x_2)x + x_1x_2) = 0 \]

Now, by the distributive property, we multiply \( a \) by each term, obtaining:

\[ ax^2 - a(x_1 + x_2)x + ax_1x_2 = 0 \]

At this point, we can compare this expression with the original equation \( ax^2 + bx + c = 0 \). In particular, we see that the coefficients must be equal. Comparing the linear term, we obtain:

\[ -a(x_1 + x_2) = b \]

Solving for \( x_1 + x_2 \), we get:

\[ x_1 + x_2 = -\frac{b}{a} \]

Similarly, comparing the constant term, we get:

\[ ax_1x_2 = c \]

Solving for the product of the roots, we get:

\[ x_1 \cdot x_2 = \frac{c}{a} \]

In summary, the roots \( x_1 \) and \( x_2 \) are related to the coefficients \( a \), \( b \), and \( c \) through these two simple relations: the sum of the roots is \( \displaystyle -\frac{b}{a} \) and the product of the roots is \( \displaystyle \frac{c}{a} \). These properties are fundamental and allow us to deduce important information about the roots without directly calculating them.

Exercise 1. Solve the quadratic equation \( x^2 - 3x - 5 = 0 \).

Solution. To solve it, we use the following formula:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

In this case, the coefficients are \( a = 1 \), \( b = -3 \), and \( c = -5 \). Applying the formula:

\[ x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(-5)}}{2(1)} = \frac{3 \pm \sqrt{9 + 20}}{2} = \frac{3 \pm \sqrt{29}}{2} \]

The solutions are:

\[ x_1 = \frac{3 + \sqrt{29}}{2} \quad , \quad  x_2 = \frac{3 - \sqrt{29}}{2} \]

Exercise 2 (reduced form). Find the solutions of the following equation \( x^2 + 6x = 0 \).

Solution. To solve it, we can factor out the common term:

\[ x(x + 6) = 0 \]

The solutions are: \( x_1 = 0 \) and \( x_2 = -6 \).

Exercise 3 (monomial equation). Find the solutions of the equation \( x^2 = 16 \).

Solution. To solve it, we can take the square root of both sides:

\[ x = \pm \sqrt{16} = \pm 4 \]

The solutions are: \( x_1 = 4 \) and \( x_2 = -4 \).

Exercise 4 (pure equation). Find the solutions of the equation \( x^2 + 9 = 0 \).

Solution. Isolate \( x^2 \):

\[ x^2 = -9 \]

Since there are no real numbers that satisfy this equation, the equation has no real solutions.

Exercise 5. Find the solutions of the following equation \( x^2 - 4 = 0 \).

Solution. Isolate \( x^2 \):

\[ x^2 = 4 \]

Now take the square root of both sides:

\[ x = \pm \sqrt{4} = \pm 2 \]

The solutions are:

\[ x_1 = 2 \quad , \quad x_2 = -2 \]

From a geometric point of view, solving a quadratic equation means finding the real values (if they exist) for which the parabola with equation \( y = ax^2 + bx + c \) intersects the x-axis, or, if you prefer, the line \( y = 0 \).