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QUADRATIC EQUATIONS 

The Quadratic Formula: For ax2 + bx + c = 0, the values of x which are the solutions of the equation are given by:

x = \dfrac{-b \pm\sqrt{b^2 – 4ac\,}}{2a}x=2ab±b24ac


For the Quadratic Formula to work, you must have your equation arranged in the form “(quadratic) = 0“. Also, the “2a” in the denominator of the Formula is underneath everything above, not just the square root. And it’s a “2a” under there, not just a plain “2“. Make sure that you are careful not to drop the square root or the “plus/minus” in the middle of your calculations, or I can guarantee that you will forget to “put them back in” on your test, and you’ll mess yourself up. Remember that “b2” means “the square of ALL of b, including its sign”, so don’t leave b2 being negative, even if b is negative, because the square of a negative is a positive.

In other words, don’t be sloppy and don’t try to take shortcuts, because it will only hurt you in the long run. Trust me on this!

Here are some examples of how the Quadratic Formula works:

  • Solve x2 + 3x – 4 = 0

This quadratic happens to factor:

x2 + 3x – 4 = (x + 4)(x – 1) = 0

…so I already know that the solutions are x= –4 and x = 1. How would my solution look in the Quadratic Formula? Using a = 1b = 3, and c = –4, my solution looks like this:

x = \dfrac{-(3) \pm \sqrt{(3)^2 – 4(1)(-4)\,}}{2(1)}x=2(1)(3)±(3)24(1)(4)

= \dfrac{-3 \pm \sqrt{9 + 16\,}}{2} = \dfrac{-3 \pm \sqrt{25\,}}{2}=23±9+16=23±25

= \dfrac{-3 \pm 5}{2} = \dfrac{-3 – 5}{2},\, \dfrac{-3 + 5}{2}=23±5=235,23+5

= \dfrac{-8}{2},\, \dfrac{2}{2} = -4,\, 1=28,22=4,1

Then, as expected, the solution is x = –4, x=1

Supose you have ax2 + bx + c = y, and you are told to plug zero in for y. The corresponding x-values are the x-intercepts of the graph. So solving ax2 + bx + c = 0 for xmeans, among other things, that you are trying to find x-intercepts. Since there were two solutions for x2 + 3x – 4 = 0, there must then be two x-intercepts on the graph. Graphing, we get the curve below:

y = x^2 + 3x - 4

As you can see, the x-intercepts (the red dots above) match the solutions, crossing the x-axis at x = –4and x = 1. This shows the connection between graphing and solving: When you are solving “(quadratic) = 0“, you are finding the x-intercepts of the graph. This can be useful if you have a graphing calculator, because you can use the Quadratic Formula (when necessary) to solve a quadratic, and then use your graphing calculator to make sure that the displayed x-intercepts have the same decimal values as do the solutions provided by the Quadratic Formula.

Note, however, that the calculator’s display of the graph will probably have some pixel-related round-off error, so you’d be checking to see that the computed and graphed values were reasonably close; don’t expect an exact match.


  • Solve 2x2 – 4x – 3 = 0. Round your answer to two decimal places, if necessary.

There are no factors of (2)(–3) = –6 that add up to –4, so I know that this quadratic cannot be factored. I will apply the Quadratic Formula. In this case, a = 2, b = –4, and c = –3:

x = \dfrac{-(-4) \pm \sqrt{(-4)^2 – 4(2)(-3)\,}}{2(2)}x=2(2)(4)±(4)24(2)(3)

= \dfrac{4 \pm \sqrt{16 + 24\,}}{4} = \dfrac{4 \pm \sqrt{40\,}}{4}=44±16+24=44±40

= \dfrac{4 \pm \sqrt{4\,} \sqrt{10\,}}{4} = \dfrac{4 \pm 2 \sqrt{10\,}}{4}=44±410=44±210

= \dfrac{2(2 \pm \sqrt{10\,})}{2(2)} = \dfrac{2 \pm \sqrt{10\,}}{2}=2(2)2(2±10)=22±10

= \dfrac{2 – \sqrt{10\,}}{2}, \dfrac{2 + \sqrt{10\,}}{2}=2210,22+10

\approx -0.58113883, 2.58113880.58113883,2.5811388

Then the answer is x = –0.58, x = 2.58, rounded to two decimal places.

Warning: The “solution” or “roots” or “zeroes” of a quadratic are usually required to be in the “exact” form of the answer. In the example above, the exact form is the one with the square roots of ten in it. You’ll need to get a calculator approximation in order to graph the x-intercepts or to simplify the final answer in a word problem. But unless you have a good reason to think that the answer is supposed to be a rounded answer, always go with the exact form.

Compare the solutions of 2x2 – 4x – 3 = 0with the x-intercepts of the graph:

y = 2x^2 - 4x - 3

Just as in the previous example, the x-intercepts match the zeroes from the Quadratic Formula. This is always true. The “solutions” of an equation are also the x-intercepts of the corresponding graph.

http://www.purplemath.com/modules/quadform.htm

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