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Algebra Tutorials - Quadratic Equations

The most general expression of a quadratic equation is shown below:

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

where \(a\), \(b\) and \(c\) are real constants, with \(a\neq 0\). For instance, the following equation:
\[2x^2 -3x + 4 = 0\]

is a quadratic equation, whereas
\[4x - 5 = 0\]

is not (because the factor \(x^2\) is not present in the equation).

Solving the Quadratic Equation

The main objective when we have a quadratic equation is to find its solutions or roots, the other name that is commonly used. The roots are computed with the well known quadratic formula

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

Example: Find the roots of the equation

\[2x^2 - x -1 = 0\]

Solution: We need to apply the quadratic equation formula, and replace the corresponding values of \(a\), \(b\) and \(c\). In this case, \(a=2\), \(b = -1\) and \(c = -1\):

\[x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} = \frac{-(-1) \pm \sqrt{(-1)^2-4\cdot 2 \cdot (-1)}}{2\cdot 2}\]

\[= \frac{-b \pm \sqrt{b^2-4ac}}{2a} = \frac{1 \pm \sqrt{1+8 }}{4} = \frac{1 \pm 3 }{4}\]

Now, we see that we have two solutions because of the \(\pm\), which means that the roots are
\[x_1 = \frac{1 + 3 }{4} = 1\]

\[x_2 = \frac{1 - 3 }{4} = -\frac{1}{2}\]

The discriminant

It turns out that we can know a lot about the roots of a quadratic equation before even solving it. How is that possible? Well, we need to compute the following quantity, which is called the Discriminant:

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

The discriminant can be negative, zero or positive, and the type of solutions will depend on it. In fact, we have that

  • If \(D > 0\): There are two different real roots

  • If \(D = 0\): There only one real root (the roots are repeated)

  • If \(D < 0\): There are no real roots (The roots are complex)

So, depending on the value of the discriminate we'll be able to determine beforehand what kind of solutions.

Why we get complex roots with a negative discriminate? Well, because in the quadratic formula, the term \( \sqrt{ b^2-4ac}\) appears, which won't be real if \(b^2-4ac <0\). To see graphically how to locate the roots, you could try the quadratic equation solver.

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