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Issue No 34, 3 July 2023
By: Anthony O. Ives
Quadratic equations are 2nd order polynomial equations which can be used to solve higher order polynomial equations. In a future article general polynomial equations will be discussed in more detail. Quadratic equations have a wide range of applications and can be used to solve various mathematical problems with engineering applications. Quadratic equations have a formulae which allows you to find their roots precisely. A root of a polynomial is the value of the independent variable (the x variable) for which the dependent variable (the y varible) will be zero.
A quadratic equation is a 2nd order polynomial equation which means that the highest index of the independent varible or x varible is 2. A quadratic equation can be expressed as two linear equations [1] multiplied together as below:
\[y = x + 4 \]
\[y = x + 3 \]
\[y = (x + 4)(x + 3) = x^2 + 7x + 12 \]
If you have trouble understanding how to multiply linear equations together the diagram below might help you understand:
The two linear equations actually define the roots of the quadratic equation as following:
\[y = x + 4 = 0, x = -4 \]
\[y = x + 3 = 0, x = -3 \]
Therefore if you insert -4 or -3 as x into the expanded quadratic equation then for both values of x the value y will be equal to 0, see the example below:
If x = -4:
\[y = x^2 + 7x + 12 = (-4)^2 + 7(-4) + 12 = 16 + (-28) + 12 = 0\]
If x = -3:
\[y = x^2 + 7x + 12 = (-3)^2 + 7(-3) + 12 = 9 + (-21) + 12 = 0\]
From the example and in expanding the quadratic equation a pattern can also noted if you determine two numbers that add together to give the x coefficient and multiply together to give the constant coefficient then you can determine the roots very quickly see example below:
\[y = x^2 + 7x + 12 = x^2 + (3+4)x + (3\times4)\]
This pattern can be see when you expand the quadratic equation:
\[y = (x + 4)(x + 3) \]
\[y = x(x + 3) + 4(x + 3) \]
\[y = x^2 + 3x + 4x + 4\times3 \]
\[y = x^2 + (3+4)x + (3\times4) \]
If there is a coefficent on the x squared term you can simply divide the whole quadratic by the x squared coefficient as in the following example:
\[y = 3x^2 + 21x + 36 = 0\]
If you divide the above equation by 3 you get the following:
\[y = x^2 + 7x + 12 = 0\]
The above equation is same equation we have been using hence the roots are -3 and -4 again as before. However, for some quadratic equations it is very difficult to find the two roots using the method that has been explained so far this is where a specific equation has been derived to determine the roots directly from the coefficients. The equation is:
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
Where \(y = ax^2 + bx + c\), so using the example we have previously looked at the two roots are determined as follows:
\[x = \frac{-7 \pm \sqrt{(7)^2 - 4(1)(12)}}{2(1)} \]
\[x = \frac{-7}{2} \pm \frac{\sqrt{49 - 48}}{2} \]
\[x = \frac{-7}{2} \pm \frac{1}{2} \]
\[x = \frac{-7}{2} + \frac{1}{2} = -3 \]
\[x = \frac{-7}{2} - \frac{1}{2} = -4 \]
Some quadratic equations will have no roots (or no real roots anyhow) depending on the value of the term \(sqrt{b^2 - 4ac}\) as explained below:
\(\sqrt{b^2 - 4ac} \ge 0\) is a quadratic equation with real roots
\(\sqrt{b^2 - 4ac} < 0\) is a quadratic equation with complex or imaginary roots
Here is an example of a quadratic equation with complex roots [2]:
\[y = (x + (4-3j))(x + (4+3j))\]
\[y = x^2 + (4-3j)x + (4+3j)x + (4-3j)(4+3j)\]
\[y = x^2 + 8x + 4(4+3j)-3j(4+3j)\]
\[y = x^2 + 8x + 16+12j-12j+9\]
\[y = x^2 + 8x + 25 = 0\]
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
\[x = \frac{-8 \pm \sqrt{(8)^2 - 4(1)(25)}}{2(1)} \]
\[x = \frac{-8}{2} \pm \frac{\sqrt{64 - 100}}{2} \]
\[x = \frac{-8}{2} \pm \frac{\sqrt{-36}}{2} \]
\[x = 4 \pm \sqrt{-1} \]
\[x = 4 + 3j, x = 4 - 3j \]
Complex roots come as conjugate pairs which means that the magnitude of the real and imaginary numbers are the same but the sign of the imaginery number is different for each root. Quadratic equations with complex numbers have important meanings for control theory and vibration theory.
It might be helpful to understand quadratic equations by drawing them in the same way as was done in reference [1], you find that the curves for quadratic equations with real roots cross the x axis at two different points where as a quadratic equation with complex roots will have a similar shape but will not cross the x axis at any point, the curve will generally be away from the x axis. The curves generated by a quadratic equation are known as parabolic curves or parabola, they are always U shaped. Parabola curves can also be used to describe specific geometry shapes.
Further information on quadratic equations and parabolic curves can be found in reference [3], in future articles polynomial equations will discussed in more detail as well as their applications.
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References:
[1] http://www.eiteog.com/EiteogBLOG/No15EiteogBlogLinear.html
[2] http://www.eiteog.com/EiteogBLOG/No31EiteogBlogNumbers.html
[3] Engineering Mathematics, K. A. Stroud, Fourth Edition, 1995, Macmillan
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