Quadratic equations. Do you need a delta to count them?

Question

Tomasz Modrzejewski · Answer

The answers probably can guess, since I write about the work of the delta in general is not necessary to calculate the roots of a quadratic equation. So to solve this equation could even junior high school student grades 1-3 who does not know this concept at all, and knows the basic behold shortcut formulas:

$1) \ \ a^{2}-b^{2}=(a+b)(a-b) \\2) \ \ a^{2} \underline{+}2ab+b^{2}=(a \underline{+} b)^{2}$

Probably most associate these simple models. However, the most important skill is to use them in the examples. I write from easiest to hardest examples and conclude on the search for complex roots, and not once did I write stamp $\Delta \ :)$

Case: 1 b = 0. Each person probably would amount to the equation of this type without using the delta, because they are very simple and there is not really what to translate.Here are a few examples:

$x^{2}-9=0 \\ (x-3)(x+3)=0 \\ x_{1}=3 \\ x_{2}=-3 \\ \hbox{LUB:} \\ x^{2}-9=0 \\ x^{2}=9 \quad /\sqrt{} \\ |x|=3 \\ x_{1}=3 \\ x_{2}=-3$

In the first process (and less timetabled) used a truncated multiplication formula 1) that I wrote. I'll do one more example, because here there is too much what to elaborate:

$x^{2}-3=0 \\ (x-\sqrt{3})(x+\sqrt{3})=0 \\ x_{1}=\sqrt{3} \\ x_{2}=-\sqrt{3}$

Case2: a = 0 Some people use the delta to count such cases, however, it is also not necessary at all. The algorithm is simple: we issue x front bracket and look forward to what x resets the second bracket. Let's make a simple two examples:

$3x^{2}+x=0 \\ x(3x+1)=0 \\3x+1=0 \\3x=-1 \quad /:3 \\ x_{1}=-\frac{1}{3} \\ x_{2}=0$

It should slide to note that in this case always being one of the elements is zero. There is no shred what happened here, we'll Example2:

$2x^{2}-7x=0 \\ x(2x-7)=0 \\2x-7=0 \\2x=7 \quad /:2 \\ x_{1}=\frac{7}{2} \\ x_{2}=0$

Case3: all the coefficients a, b, c are nonzero And here we should now turn thinking. I'll start with simple examples or "foldable" to model condensed multiplication (model2)). Example1:

$x^{2}+2x+1=0$

This is a basic example. Note that the square root of the coefficient c and the square root of the ratio is1 and their product, and two is2, ie, the coefficient b. Thus, we can easily collapse and transform as follows:

$(x+1)^{2}=0 \\ x+1=0 \\ x=-1$

Another example of this type:

$x^{2}-6x+9=0$

Here we see the same way that the root of the c =3 and root for =1 and their product is multiplied by2 equals 6 where once we have:

$(x-3)^{2}=0 \\ x-3=0 \\ x=3$

They can also be fractions. For example:

$x^{2}+\frac{2}{3}x+\frac{1}{9}=0 \\ \left( x+\frac{1}{3} \right)^{2}=0 \\ x+\frac{1}{3}=0 \\ x=-\frac{1}{3}$

$4x^{2}+4x+1=0 \\ (2x+1)^{2}=0 \\2x+1=0 \\2x=-1 \\ x=-\frac{1}{2}$

Yes, but not all, after all roll up in this pattern. What then? Then we will use both designs. And all will be to the separation factor "c" on the two figures. Let's do an example:

$x^{2}-4x+3=0$

In this example, certainly everyone (or at least most) would benefit from the delta. But rozpiszmy three as 4-1 and note the following thing:

$x^{2}-4x+4-1=0$

Note that

$x^{2}-4x+4=(x-2)^{2}$

Thus we have

$(x-2)^{2}-1=0$

We use the formula

$(x-2-1)(x-2+1)=0 \\ (x-3)(x-1)=0 \\ \hbox{I od razu mamy pierwiastki:} \\ x_{1}=3 \\ x_{2}=1$

Note: your wits can not know borders.

Now I will give you another example:

$x^{2}+x-2=0$

Here we split x to 2x-x and obtain

$x^{2}+2x-x-2=0$

Taught in the first two components of x before the parenthesis:

$x(x+2)-(x+2)=0$

Taught factor (x +2) before the parenthesis:

$(x-1)(x+2)=0 \\ x_{1}=1 \\ x_{2}=-2$

Now some examples of consolidation:

$x^{2}-5x-6=0 \\ x^{2}-6x+x-6=0 \\ x(x-6)+(x-6)=0 \\ (x+1)(x-6)=0 \\ x_{1}=-1 \\ x_{2}=6 \\ \\2x^{2}+7x+3=0 \\2x^{2}+6x+x+3=0 \\2x(x+3)+(x+3)=0 \\ (2x+1)(x+3)=0 \\2x+1=0 \\2x=-1 \\ x_{1}=-\frac{1}{2} \\ x_{2}=-3$

Well, we received a nice measurable elements. Received some irrational:

$x^{2}+4x-1=0 \\ x^{2}+4x+2-5=0 \\ (x+2)^{2}-5=0 \\ (x+2-\sqrt{5})(x+2+\sqrt{5})=0 \\ \hbox{Skad:} \\ x_{1}=\sqrt{5}-2 \\ x_{2}=-2-\sqrt{5}$

$2x^{2}+4x-7=0 \quad /:2\\ x^{2}+2x-\frac{7}{2}=0 \\ x^{2}+2x+1-\frac{9}{2}=0 \\ (x+1)^{2}-\frac{9}{2}=0 \\ (x+1+\frac{3}{\sqrt{2}})(x+1-\frac{3}{\sqrt{2}})=0 \\ x_{1}=-1-\frac{3}{\sqrt{2}}=-1-\frac{3\sqrt{2}}{2} \\ x_{2}= \frac{3\sqrt{2}}{2}-1$ $3x^{2}+5x-1=0 \quad /:3 \\ x^{2}+\frac{5}{3}x-\frac{1}{3}=0 \\ x^{2}+\frac{5}{3}x+\frac{25}{36}-\frac{37}{36}=0 \\ (x+\frac{5}{6})^{2}-\frac{37}{36}=0 \\ (x+\frac{5}{6}-\frac{\sqrt{37}}{6})(x+\frac{5}{6}+\frac{\sqrt{37}}{6})=0 \\ x_{1}= \frac{\sqrt{37}-5}{6} \\ x_{2}=\frac{-5-\sqrt{37}}{6}$

And a last one:

$5x^{2}+2x-9=0 \quad /:5 \\ x^{2}+\frac{2}{5}x-\frac{9}{5}=0 \\ x^{2}+\frac{2}{5}x+\frac{4}{100}-\frac{184}{100}=0 \\ (x+\frac{2}{10})^{2}-\frac{184}{100}=0 \\ (x+\frac{2}{10}+\frac{\sqrt{184}}{10})(x+\frac{2}{10}-\frac{\sqrt{184}}{10})=0 \\ x_{1}= \frac{-2-\sqrt{184}}{10}=\frac{-2-2\sqrt{46}}{10}=\frac{-1-\sqrt{46}}{5} \\ x_{2}=\frac{\sqrt{184}-2}{10}=\frac{\sqrt{46}-1}{5}$

As you can see from all the examples, starting with two models truncated multiplication, you can come to a solution. Now, for the curious: How to calculate the elements without using the delta when this delta is really negative?

Case1: b = 0

$x^{2}+16=0$

Proceed as in that case. Except that

$(4i)^{2}=-16$ .

therefore

$x^{2}-(4i)^{2}=0 \\ (x-4i)(x+4i)=0 \\ x_{1}=4i \\ x_{2}=-4i$

Here, too, there is too much philosophy, so even just one example:

Note: When c = 0 ALWAYS delta is positive.

$x^{2}+7=0 \\ (x-i\sqrt{7})(x+i\sqrt{7})=0 \\ x_{1}=i\sqrt{7} \\ x_{2}=-i\sqrt{7}$

Case2: all the coefficients are non-zero. We act basically as it would be real roots. Here are a few examples:

$x^{2}+2x+6=0 \\ x^{2}+2x+1+5=0 \\ (x+1)^{2}+5=0 \\ (x+1)^{2}-(i\sqrt{5})^{2}=0 \\ (x+1+i\sqrt{5})(x+1-i\sqrt{5})=0 \\ x_{1}=-(1+i\sqrt{5}) \\ x_{2}=i\sqrt{5}-1$

$2x^{2}+6x+9=0 \quad /:2 \\ x^{2}+3x+\frac{9}{2}=0 \\ x^{2}+3x+\frac{9}{4}+\frac{9}{4}=0 \\ (x+\frac{3}{2})^{2}+\frac{9}{4}=0 \\ (x+\frac{3}{2})^{2}-(\frac{3}{2}i)^{2}=0 \\ (x+\frac{3}{2}-\frac{3}{2}i)(x+\frac{3}{2}+\frac{3}{2}i)=0 \\ x_{1}= \frac{3}{2}(i-1) \\ x_{2}=-\frac{3}{2}(i+1)$

I'll think of more recent example of nowhere taken:

$3x^{2}-7x+12=0 \quad /:3 \\ x^{2}-\frac{7}{3}+4=0 \\ x^{2}-\frac{7}{3}+\frac{49}{36}+\frac{95}{36}=0 \\ (x-\frac{7}{6})^{2}+\frac{95}{36}=0 \\ (x-\frac{7}{6})^{2}-(\frac{\sqrt{95}}{6}i)^{2}=0 \\ (x-\frac{7}{6}+\frac{\sqrt{95}i}{6})(x-\frac{7}{6}-\frac{\sqrt{95}i}{6})=0 \\ x_{1}= \frac{7-i\sqrt{95}}{6} \\ x_{2}=\frac{7+i\sqrt{95}}{6}$

I ONLY use two formulas can be calculated multiplication condensed elements of all quadratic equations. And how do you think that did these designs on x1, x2? Only through these two fundamental rights, which I wrote at the beginning :) For the ignorant:

$\sqrt{-x}=i\sqrt{x}$

So, for example,

$\sqrt{-9}=i\sqrt{9}=3i$

a i is the imaginary unit, i.e.

$i^{2}=-1$

I ... I do not currently using the delta :) And do you think that where did the designs for x1 and x2? This is all thanks to the formulas 1) and2). To prove it to you, bring out these patterns :)

$ax^{2}+bx+c=0 \qquad /:a \\ x^{2}+\frac{b}{a}x+\frac{c}{a}=0 \\ x^{2}+\frac{b}{a}x+ \left( \frac{b}{2a} \right)^{2} +\left(\frac{c}{a}-\frac{b^{2}}{4a^{2}}\right) =0 \\ \left( x+\frac{b}{2a} \right)^{2} - \left(\frac{b^{2}-4ac}{4a^{2}} \right)=0 \\ \left( x+\frac{b}{2a}-\sqrt{\frac{b^{2}-4ac}{4a^{2}} \right) \left(x+\frac{b}{2a}- \sqrt{\frac{b^{2}-4ac}{4a^{2}} \right)=0$

Comment out now our b ^ 2-4ac on delta finally get:

$\left( x+\frac{b}{2a}+\frac{\sqrt{\Delta}}{2a} \right) \left( x+\frac{b}{2a}-\frac{\sqrt{\Delta}}{2a} \right)=0 \\ x_{1}=-\frac{b}{2a}-\frac{\sqrt{\Delta}}{2a}=\frac{-b-\sqrt{\Delta}}{2a} \\ x_{2}=-\frac{b}{2a}+\frac{\sqrt{\Delta}}{2a}=\frac{-b+\sqrt{\Delta}}{2a}$

Hmm ... I promised that I would not use the Greek letters Delta, but I used ... Well, but ... :) outputting these patterns showed his counting algorithm to understand it more.

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Quadratic equations. Do you need a delta to count them?

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