Quadratic Equations: Very Difficult Problems with Solutions

Solution:
The equation is defined for x, such that [tex]x-2 \ne 0; x+2 \ne 0; x^2-4 \ne 0[/tex], which yield us [tex]x \ne \pm 2[/tex].
[tex]\frac{5}{2-x}+\frac{x-5}{x+2}+\frac{3x+8}{x^2-4}=0[/tex]
[tex]\frac{5(x+2)}{(2-x)(x+2)}+\frac{(x-5)(2-x)}{(x+2)(2-x)}+\frac{3x+8}{x^2-4}=0[/tex]
Have in mind that [tex]\frac{3x+8}{x^2 - 4}=-\frac{3x+8}{4-x^2}[/tex]
[tex]\frac{5x+10}{4-x^2}+\frac{2x-10-x^2+5x}{4-x^2}-\frac{3x+8}{4-x^2}=0[/tex]. We don't need the denominator anymore.
[tex]5x+10+7x-10-x^2-3x-8=0[/tex]
[tex]x^2-9x+8=0[/tex]
[tex]x^2-8x-x+8=0[/tex]
[tex]x(x-8)-(x-8)=0[/tex]
[tex](x-1)(x-8)=0[/tex], so the roots are x=1 and x=8. The equation is defined for them, so they are both solutions.

Solution:
a is obviously a non-zero number.
Let divide the equation by [tex]a^2[/tex].
So we get [tex]\frac{x^2}{a^2}-2\frac{ax}{a^2}+\frac{a^2}{a^2}=0[/tex]
[tex](\frac{x}{a})^2-2.\frac{x}{a}+1=0[/tex], or [tex](\frac{x}{a}-1)^2=0[/tex], which yields [tex]\frac{x}{a}=1[/tex].

Solution:
[tex]\frac{x^{2002} + 4x^{2001}}{4x^{2000}} = 2449.25[/tex]
Multiply the both sides by 4, we get,
[tex]\frac{x^{2002} + 4x^{2001}}{x^{2000}} = 9797[/tex]

[tex]\frac{x^{2000} (x^2 + 4x)}{x^{2000}} = 9797[/tex]
[tex]x^2 + 4x =9797[/tex]
[tex]x^2 + 4x - 9797=0[/tex]
[tex]x^2 + 101x - 97x -9797 = 0[/tex]
[tex]x(x+101) -97 (x+101) = 0[/tex]
[tex](x+101)(x-97) = 0[/tex]
The roots of the equation are -101 and 97.

Solution:
Firstly, lets say that the first page is n.
Hence, it is clear that the next one is n+1.
The product of the pages is equal to 1122, so
n * (n + 1) = 1122
We multiply using the distributive property, and then the equation becomes [tex]n^2 + n = 1122[/tex].
We move everything to the left, making the equation equal to zero [tex]n^2 + n – 1122 = 0[/tex].
Eventually, we factorize the second degree equation, and we get [tex](n + 34)(n – 33) = 0[/tex].
Since the product of the two "numbers" is equal to zero, then one of them is equal to zero.
Let's take the first one.
[tex]n + 34 = 0[/tex].
Then [tex]n = - 34[/tex], which is impossible, since a page can't have a negative number. Let's take the second one.
[tex]n - 33 = 0[/tex].
Then, [tex]n = 33[/tex], which is accepted.
So, the first page is page 33, and the second is page n + 1 = 33 + 1 = page 34.

Solution:
Let [tex]x[/tex] be [tex]\sqrt{6+(\sqrt{6+(\sqrt{6...}}}[/tex]
[tex]x = \sqrt{6+(\sqrt{6+(\sqrt{6...}}}[/tex]
[tex]x^2 = 6 + \sqrt{6+(\sqrt{6+(\sqrt{6...}}}[/tex]
[tex]\Rightarrow x^2=6+x[/tex]
[tex]x^2- x - 6 = 0[/tex]
[tex](x+2)(x-3) = 0[/tex]
[tex]x = -2[/tex] or [tex]x = 3[/tex]
[tex]x[/tex] must be greater than 0, so

[tex]\sqrt{6+(\sqrt{6+(\sqrt{6...}}} = 3[/tex]

Solution:
Firstly, let us assume that the digit we are looking for is
x
Then, the whole number becomes
[tex](xx)^{-}[/tex]
Whose value is equal to
[tex](xx)^{-}=10x+x=11x[/tex]
Subsequently, we raise the number to the second power, and becomes
[tex](xx)^{-}=(11x)^2=121x^2[/tex]
Now that our number is raised to the second power, we proceed to the second part. If we assume that our original number is
[tex](xx)^{-}[/tex]
It is clear that the number produced after the squaring is equal to
[tex]((x-1)(x-1)\frac{x}{2}\frac{x}{2} )^{-}[/tex]
And its value is
[tex]1000(x-1)+100(x-1)+10\frac{x}{2}+1\frac{x}{2}[/tex]
By applying the distributive property, the number becomes
[tex]1000x-1000+100x-100+ \frac{10x}{2}+\frac{x}{2}[/tex]
To add the terms together, it is necessary to convert all amounts to like quantities. This can be achieved by multiplying each term (apart from the two last) by 2/2
[tex]\frac{1000x\cdot 2}{2}-\frac{1000\cdot 2}{2}+\frac{100x\cdot 2}{2}-\frac{100\cdot 2}{2}+\frac{10x}{2}+\frac{x}{2}[/tex]
After, the number becomes
[tex]\frac{2000x}{2}-\frac{2000}{2}+\frac{200x}{2}-\frac{200}{2}+\frac{10x}{2}+\frac{x}{2}[/tex]
We add the terms together, making the number
[tex]\frac{2000x-2000+200x-200+10x+x}{2}[/tex]

Adding the numerator’s terms we get
[tex]\frac{2211x-2200}{2}[/tex]
Now, after finishing with both our numbers, we equate them
[tex]121x^2=\frac{2211x-2200}{2}[/tex]
We multiply each side by 2, and the equation becomes
[tex]242x^2=2211x-2200[/tex]
We move all terms to the left, making the second degree trinomial equal to zero
[tex]242x^2-2211x+2200=0[/tex]
After, we can factor our expression, just to make things easier.
We notice that 242, 2211 and 2200 have a common factor of 11.
So, we can factor the whole expression into
[tex]11 (22x^2-201x+200)=0[/tex]
We divide both sides by eleven, and we end up with
[tex]22x^2-201x+200=0[/tex]
Now, it is clear that we have got a second degree equation, which is equal to zero. Our first task is to calculate the discriminant, which is given by the formula
[tex]D=b^2-4ac[/tex], where a=22, b=201 and c=200
We replace the variables with the values, and the discriminant becomes equal to
D=22801
Now, we need to figure out whether the equation has got real roots or not.
We have the following possibilities:
[tex]b^2-4ac>0\Rightarrow[/tex] There are two real roots.
The solution x of the second degree equation is given by the formula
[tex]x_{1,2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]
[tex]x_{1,2}={201\pm 151}{44}[/tex]
[tex]x_1=\frac{201+151}{44}=\frac{352}{44}=8[/tex]
[tex]x_2=\frac{201-151}{44}=\frac{50}{44}=1.13636363636...[/tex]
Finally, we come to the point where we have to choose between the two possible solutions for x. Since the value we need concerns a digit, it has to be an integer.
[tex]8 \in Z [/tex]   True
[tex]1.13636363636... \in Z[/tex]   False

Hence, we come to the conclusion that x=8.
[tex](xx)^{-}=88[/tex]