Quadratic Equations: Very Difficult Problems with Solutions

Solution:
[tex]a[/tex] 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 \times \frac{x}{a}+1=0[/tex], or [tex](\frac{x}{a}-1)^2=0[/tex], which yields [tex]\frac{x}{a}=1[/tex].

Solution:
The equation is defined for [tex]x[/tex], 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 [tex]x=1[/tex] and [tex]x=8[/tex]. The equation is defined for them, so they are both solutions.

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:
Firstly, let us assume that the digit we are looking for is
x
Then, the whole number becomes
[tex]\overline{xx}[/tex]
Whose value is equal to
[tex]\overline{xx}=10x+x=11x[/tex]
Subsequently, we raise the number to the second power, and becomes
[tex]\overline{xx}^{\,2}=(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]\overline{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 [tex]\frac{2}{2}[/tex]
[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]\overline{xx}=88[/tex]

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 or x = 3[/tex]
[tex]x[/tex] must be greater than 0, so

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

Solution:
Pair the factors so that the sums of the numbers are equal ([tex]3+8=4+7=11[/tex]):
[tex](x-3)(x-8)=x^2-11x+24[/tex], [tex](x-4)(x-7)=x^2-11x+28[/tex].
Let [tex]t=x^2-11x+24[/tex]: [tex]t(t+4)=60[/tex], [tex]t^2+4t-60=0[/tex], [tex]t_1=6[/tex], [tex]t_2=-10[/tex].
[tex]x^2-11x+24=6[/tex]: [tex]x^2-11x+18=0[/tex], [tex]x=2[/tex] or [tex]x=9[/tex].
[tex]x^2-11x+24=-10[/tex]: [tex]x^2-11x+34=0[/tex], [tex]D=121-136<0[/tex], no real roots.
Answer: [tex]x_1=2[/tex], [tex]x_2=9[/tex].

Solution:
Let [tex]f(x)=x^2-2ax+a+2[/tex]; its graph is a parabola opening upward with vertex at [tex]x=a[/tex]. Both roots lie in [tex](1, 3)[/tex] exactly when:
[tex]\frac{D}{4}=a^2-a-2=(a-2)(a+1)\ge 0[/tex]: [tex]a\le -1[/tex] or [tex]a\ge 2[/tex];
the vertex is inside the interval: [tex]1\lt a<3[/tex];
[tex]f(1)=3-a>0[/tex]: [tex]a<3[/tex];
[tex]f(3)=11-5a>0[/tex]: [tex]a\lt \frac{11}{5}[/tex].
All four hold for [tex]2\le a\lt \frac{11}{5}[/tex], i.e. [tex]a\in[2, \frac{11}{5})[/tex]. (For [tex]a=2[/tex] there is a double root [tex]x=2[/tex].)

Solution:
[tex]f(x)=0[/tex] gives [tex]x^2-ax+3a-9=0[/tex], i.e. [tex](x-3)(x-a+3)=0[/tex]. So [tex]A[/tex] and [tex]B[/tex] have [tex]x=3[/tex] and [tex]x=a-3[/tex], and [tex]AB=|a-6|[/tex] (we need [tex]a\ne 6[/tex]).
[tex]C=(0, 9-3a)[/tex], so the height to [tex]AB[/tex] is [tex]|9-3a|=3|a-3|[/tex].
Area: [tex]\frac{1}{2}|a-6|\times 3|a-3|=27[/tex], i.e. [tex]|(a-6)(a-3)|=18[/tex].
[tex](a-6)(a-3)=18[/tex]: [tex]a^2-9a=0[/tex], [tex]a=0[/tex] or [tex]a=9[/tex].
[tex](a-6)(a-3)=-18[/tex]: [tex]a^2-9a+36=0[/tex], [tex]D<0[/tex], no solutions.
Answer: [tex]a=0[/tex] or [tex]a=9[/tex].

Solution:
If [tex]a=0[/tex], the root is [tex]x=\frac{9}{4}[/tex], not an integer.
Let [tex]x_0[/tex] be an integer root. [tex]x_0\ne 0[/tex] (otherwise [tex]9=0[/tex]), and [tex]ax_0^2=4x_0-9[/tex].
The left side is divisible by [tex]x_0[/tex], so [tex]x_0[/tex] divides [tex]4x_0-9[/tex], hence [tex]x_0[/tex] divides [tex]9[/tex]: [tex]x_0\in\{\pm 1, \pm 3, \pm 9\}[/tex].
[tex]a=\frac{4x_0-9}{x_0^2}[/tex]: for [tex]x_0=1[/tex], [tex]a=-5[/tex]; for [tex]x_0=-1[/tex], [tex]a=-13[/tex]; for [tex]x_0=3[/tex], [tex]-3[/tex], [tex]9[/tex], [tex]-9[/tex] we get [tex]\frac{1}{3}[/tex], [tex]-\frac{7}{3}[/tex], [tex]\frac{1}{3}[/tex], [tex]-\frac{5}{9}[/tex] - not integers.
Answer: [tex]a=-5[/tex] (root [tex]1[/tex]) or [tex]a=-13[/tex] (root [tex]-1[/tex]).

Solution:
Expand: [tex]x^2+2xy+y^2=xy-x+y-1[/tex], i.e. [tex]x^2+(y+1)x+y^2-y+1=0[/tex].
Treat it as a quadratic equation in [tex]x[/tex]. It has a real solution only if its discriminant is non-negative:
[tex]D=(y+1)^2-4(y^2-y+1)=-3y^2+6y-3=-3(y-1)^2\ge 0[/tex].
This holds only for [tex]y=1[/tex]. Then [tex]D=0[/tex] and [tex]x=-\frac{y+1}{2}=-1[/tex].
The only solution is [tex]x=-1[/tex], [tex]y=1[/tex] (check: [tex]0^2=0\times 0[/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:
Since [tex]a[/tex], [tex]b[/tex], [tex]c[/tex] are not all equal, they are not all zero, so [tex]a^2+b^2+c^2>0[/tex] and the equation is quadratic.
[tex]\frac{D}{4}=(a+b+c)^2-3(a^2+b^2+c^2)=2ab+2bc+2ca-2a^2-2b^2-2c^2[/tex]
[tex]=-\left[(a-b)^2+(b-c)^2+(c-a)^2\right][/tex].
The sum of the three squares is [tex]0[/tex] only when [tex]a=b=c[/tex]. Here they are not all equal, so the sum is positive and [tex]D<0[/tex].
Hence the equation has no real roots.

Solution:
Since [tex]b>0[/tex], the equation is quadratic. Factor the discriminant twice with the difference of squares:
[tex]D=(b^2+c^2-a^2)^2-(2bc)^2=(b^2+c^2-a^2-2bc)(b^2+c^2-a^2+2bc)[/tex]
[tex]=\left[(b-c)^2-a^2\right]\left[(b+c)^2-a^2\right]=(b-c-a)(b-c+a)(b+c-a)(b+c+a)[/tex].
By the triangle inequality: [tex]b\lt c+a[/tex], so [tex]b-c-a<0[/tex]; [tex]c\lt a+b[/tex], so [tex]b-c+a>0[/tex]; [tex]a\lt b+c[/tex], so [tex]b+c-a>0[/tex]; and [tex]b+c+a>0[/tex].
Exactly one factor is negative, so [tex]D<0[/tex] and the equation has no real roots.

Solution:
Suppose the equation has a rational root [tex]\frac{p}{q}[/tex] written in lowest terms ([tex]p[/tex] and [tex]q[/tex] integers, not both even). Multiply [tex]a\frac{p^2}{q^2}+b\frac{p}{q}+c=0[/tex] by [tex]q^2[/tex]:
[tex]ap^2+bpq+cq^2=0[/tex].
If [tex]p[/tex] and [tex]q[/tex] are both odd, all three terms are odd, and the sum of three odd numbers is odd, not [tex]0[/tex].
If exactly one of [tex]p[/tex], [tex]q[/tex] is even, exactly two of the terms are even and the third is odd, so the sum is again odd, not [tex]0[/tex].
Both cases give a contradiction, so the equation has no rational roots.

Solution:
Let [tex]f(x)=ax^2+bx+c=a(x-x_1)(x-x_2)[/tex] with [tex]0\lt x_1\lt x_2<1[/tex].
[tex]f(0)=c=ax_1x_2>0[/tex] and [tex]f(1)=a+b+c=a(1-x_1)(1-x_2)>0[/tex] are positive integers, so each is at least [tex]1[/tex] and
[tex]1\le f(0)f(1)=a^2\,x_1(1-x_1)\,x_2(1-x_2)[/tex].
For [tex]0\lt x<1[/tex] we have [tex]x(1-x)=\frac{1}{4}-\left(x-\frac{1}{2}\right)^2\le\frac{1}{4}[/tex], with equality only at [tex]x=\frac{1}{2}[/tex]. The roots are distinct, so they cannot both equal [tex]\frac{1}{2}[/tex], and [tex]1\lt \frac{a^2}{16}[/tex], i.e. [tex]a>4[/tex].
For [tex]a=5[/tex] it works: [tex]5x^2-5x+1=0[/tex] has roots [tex]\frac{5\pm\sqrt{5}}{10}[/tex], both in [tex](0, 1)[/tex].
Answer: [tex]a=5[/tex].