Quadratic Equations: Difficult Problems with Solutions

Solution:
[tex]x^2-9x+20=x^2-4x-5x+20=x(x-4)-5(x-4)=(x-5)(x-4).[/tex]
The solutions to the equation are 5 and 4.
Their difference is 1.

Solution:
By the short formula for the discriminant, it is [tex]10^2+69=100+69=169=13^2[/tex]. The roots are [tex]x_{1,2}=10 \pm 13[/tex], or [tex]x_1 = 23[/tex] and [tex]x_2=-3[/tex].

Solution:
$x^2-6x+9 = (x-3)^2$

$(x-3)^2 = x-3$

Let $x-3 = y$

$y^2=y$

$y^2-y=0$

$y(y-1)=0$

$y_1=0, y_2= 1$

$x=y+3 \Rightarrow$

$x_1=3,\ x_2= 4$

Solution:
[tex]x^2-4x+4+2x=7x-14[/tex]
[tex]x^2-9x+18=0[/tex]. The discriminant is [tex]D=9^2-4 \times 18=9(9-4 \times 2)=9 \times 1=3^2[/tex]. Then the roots are [tex]x_{1,2}=\frac{9 \pm 3}{2}[/tex] or [tex]x_1=\frac{9+3}{2}=6[/tex] and [tex]x_2=\frac{9-3}{2}=3[/tex].

Solution:
[tex]\frac{6x}{10}=\frac{x^2+8}{10}[/tex]
[tex]x^2+8=6x[/tex]
[tex]x^2-6x+8=0[/tex]
[tex]x^2-4x - 2x+8=0[/tex]
[tex]x(x-4)-2(x-4)=0[/tex]
[tex](x-2)(x-4)=0[/tex]

So the roots are [tex]x_1=2[/tex] and [tex]x_2=4[/tex].

Solution:
[tex]x[/tex] must not be zero for the equation to be defined. We cross multiply and get [tex]2(x+8)=x(x+2)[/tex], or [tex]2x+16=x^2+2x[/tex], which is equivalent to [tex]x^2=16[/tex]. Then [tex]x_1=-4[/tex] and [tex]x_2=4[/tex].

Solution:
[tex]154 +11x +14x +x^2 = 304 x^2 +25x +154=304 x^2+25x+154-304=0 x^2+25x-150=0 x^2+30x-5x-150=0 x(x+30) - 5(x+30)=0 (x-5) (x+30) = 0[/tex]
Therfore, [tex]x-5=0 \Leftrightarrow x=5[/tex]
[tex]x+30=0 \Leftrightarrow x=-30[/tex]

Solution:
[tex]x^4-3x^2-4=x^4-4x^2+x^2-4=x^2(x^2-4)+(x^2-4)=(x^2+1)(x^2-4)[/tex]. Therefore the roots to the equation are [tex]x_1=-2[/tex] and [tex]x_2=2[/tex].

Solution:
Let the numbers be [tex]n[/tex] and [tex]n+1[/tex]. Then
[tex]n(n+1)=n^2+(n+1)^2-57[/tex]
[tex]n^2+n=2n^2+2n+1-57[/tex]
[tex]n^2+n-56=0[/tex], [tex]D=1+224=225[/tex], [tex]n_{1,2}=\frac{-1\pm 15}{2}[/tex], [tex]n_1=7[/tex], [tex]n_2=-8[/tex].
[tex]-8[/tex] is not a natural number, so the numbers are [tex]7[/tex] and [tex]8[/tex].
Check: [tex]49+64-56=57[/tex].

Solution:
Let the number be [tex]x[/tex]. Then [tex]x+\frac{1}{x}=4.25=\frac{17}{4}[/tex].
Multiply by [tex]4x\ne 0[/tex]: [tex]4x^2+4=17x[/tex], i.e. [tex]4x^2-17x+4=0[/tex].
[tex]D=289-64=225[/tex], [tex]x_{1,2}=\frac{17\pm 15}{8}[/tex], [tex]x_1=4[/tex], [tex]x_2=\frac{1}{4}[/tex].
Only [tex]4[/tex] is a natural number: [tex]4+\frac{1}{4}=4.25[/tex].

Solution:
An [tex]n[/tex]-gon has [tex]\frac{n(n-3)}{2}[/tex] diagonals: from each vertex there are [tex]n-3[/tex] diagonals, and each diagonal is counted from both of its ends.
[tex]\frac{n(n-3)}{2}=n+12[/tex], [tex]n^2-3n=2n+24[/tex], [tex]n^2-5n-24=0[/tex].
[tex]D=25+96=121[/tex], [tex]n_{1,2}=\frac{5\pm 11}{2}[/tex], [tex]n_1=8[/tex], [tex]n_2=-3[/tex] (rejected).
It is an octagon: [tex]8[/tex] sides and [tex]20[/tex] diagonals.

Solution:
Let the shorter leg be [tex]x[/tex] inches; the other is [tex]x+8[/tex]. The area of a right triangle is half the product of its legs:
[tex]\frac{x(x+8)}{2}=154[/tex], [tex]x^2+8x-308=0[/tex].
[tex]D=64+1232=1296=36^2[/tex], [tex]x_{1,2}=\frac{-8\pm 36}{2}[/tex], [tex]x_1=14[/tex], [tex]x_2=-22[/tex] (rejected).
The legs are [tex]14[/tex] and [tex]22[/tex] inches.

Solution:
Let the numbers be [tex]x[/tex], [tex]x+2[/tex], [tex]x+4[/tex]. Then
[tex]x^2+(x+2)^2=(x+4)^2[/tex]
[tex]2x^2+4x+4=x^2+8x+16[/tex]
[tex]x^2-4x-12=0[/tex], [tex](x-6)(x+2)=0[/tex], [tex]x=6[/tex] or [tex]x=-2[/tex].
[tex]-2[/tex] is not natural, so the numbers are [tex]6[/tex], [tex]8[/tex], [tex]10[/tex]: [tex]36+64=100[/tex].

Solution:
[tex]D=(7\sqrt{3})^2-4\times 6\times(-15)=147+360=507=169\times 3[/tex], so [tex]\sqrt{D}=13\sqrt{3}[/tex].
[tex]x_{1,2}=\frac{-7\sqrt{3}\pm 13\sqrt{3}}{12}[/tex]
[tex]x_1=\frac{6\sqrt{3}}{12}=\frac{\sqrt{3}}{2}[/tex], [tex]x_2=\frac{-20\sqrt{3}}{12}=-\frac{5\sqrt{3}}{3}[/tex].

Solution:
[tex]D=9a^2+16a^2=25a^2[/tex], so [tex]\sqrt{D}=5|a|[/tex] and [tex]x_{1,2}=\frac{-3a\pm 5a}{2}[/tex] (the sign of [tex]a[/tex] does not matter because of [tex]\pm[/tex]).
[tex]x_1=a[/tex], [tex]x_2=-4a[/tex].
Check by factoring: [tex](x-a)(x+4a)=x^2+3ax-4a^2[/tex].
For [tex]a=0[/tex] both roots coincide: [tex]x=0[/tex].

Solution:
Group the terms in pairs:
[tex]x^2(x-3)-4(x-3)=(x-3)(x^2-4)=(x-3)(x-2)(x+2)=0[/tex].
The roots are [tex]x_1=3[/tex], [tex]x_2=2[/tex], [tex]x_3=-2[/tex].

Solution:
Replace [tex]x[/tex] in the formula: [tex]f(2x)=2x-3[/tex], [tex]f(5)=5-3=2[/tex].
[tex](x-3)(2x-3)=2[/tex], [tex]2x^2-9x+9=2[/tex], [tex]2x^2-9x+7=0[/tex].
[tex]D=81-56=25[/tex], [tex]x_{1,2}=\frac{9\pm 5}{4}[/tex], [tex]x_1=\frac{7}{2}[/tex], [tex]x_2=1[/tex].

Solution:
For [tex]ax^2+bx+c=0[/tex] with [tex]a+b+c=0[/tex] we have [tex]c=-a-b[/tex], and
[tex]ax^2+bx-a-b=a(x^2-1)+b(x-1)=(x-1)(ax+a+b)[/tex].
So [tex]x_1=1[/tex] and [tex]x_2=-\frac{a+b}{a}=\frac{c}{a}[/tex].
Here [tex]2019-2000-19=0[/tex], so [tex]x_1=1[/tex], [tex]x_2=-\frac{19}{2019}[/tex].

Solution:
Substitute [tex]y=\frac{1}{2}[/tex]: [tex]\frac{a-5}{4}-\frac{a-2}{2}+a=0[/tex].
Multiply by [tex]4[/tex]: [tex]a-5-2a+4+4a=0[/tex], [tex]3a-1=0[/tex], [tex]a=\frac{1}{3}[/tex].
The equation becomes [tex]-\frac{14}{3}y^2+\frac{5}{3}y+\frac{1}{3}=0[/tex]; multiply by [tex]-3[/tex]: [tex]14y^2-5y-1=0[/tex].
[tex]D=25+56=81[/tex], [tex]y_{1,2}=\frac{5\pm 9}{28}[/tex], [tex]y_1=\frac{1}{2}[/tex], [tex]y_2=-\frac{1}{7}[/tex].

Solution:
The expression [tex]x^2-5x[/tex] appears twice, so we substitute [tex]t=x^2-5x[/tex]:
[tex]t^2-8t-84=0[/tex], [tex]D=64+336=400[/tex], [tex]t_{1,2}=\frac{8\pm 20}{2}[/tex], [tex]t_1=14[/tex], [tex]t_2=-6[/tex].
[tex]x^2-5x=14[/tex]: [tex]x^2-5x-14=0[/tex], [tex](x-7)(x+2)=0[/tex], [tex]x=7[/tex] or [tex]x=-2[/tex].
[tex]x^2-5x=-6[/tex]: [tex]x^2-5x+6=0[/tex], [tex](x-2)(x-3)=0[/tex], [tex]x=2[/tex] or [tex]x=3[/tex].
The equation has four roots: [tex]-2[/tex], [tex]2[/tex], [tex]3[/tex], [tex]7[/tex].

Solution:
Let the numbers be [tex]x[/tex] and [tex]x+1[/tex]. Then
[tex](x+1)^3-x^3=3x^2+3x+1=61[/tex], so [tex]x^2+x-20=0[/tex].
[tex](x+5)(x-4)=0[/tex], [tex]x=4[/tex] or [tex]x=-5[/tex].
Both values are integers, so there are two pairs: [tex]4[/tex], [tex]5[/tex] and [tex]-5[/tex], [tex]-4[/tex].
Check: [tex]125-64=61[/tex] and [tex](-4)^3-(-5)^3=-64+125=61[/tex].

Solution:
We cross multiply: [tex](2x-1)(2x+2)=(7x-1)(x-1)[/tex]
[tex]4x^2+4x-2x-2=7x^2-7x-x+1[/tex] [tex]4x^2+2x-2=7x^2-8x+1[/tex] [tex]3x^2-10x+3=0[/tex] The discriminant is [tex]D=10^2-4\cdot 3\cdot 3=100-36=64=8^2[/tex]. The roots are [tex]x_{1,2}=\frac{10 \pm 8}{2 \cdot 3}[/tex]. [tex]x_1=\frac{10-8}{2\cdot3}=\frac{2}{2\cdot3}=\frac{1}{3}[/tex]. [tex]x_2=\frac{10+8}{2\cdot 3}=\frac{18}{6}=3[/tex]

Solution:
We need [tex]D>0[/tex]:
[tex]D=(m+2)^2-4(5m+1)=m^2+4m+4-20m-4=m^2-16m=m(m-16)>0[/tex].
The parabola [tex]m(m-16)[/tex] is positive outside its roots [tex]0[/tex] and [tex]16[/tex].
Answer: [tex]m<0[/tex] or [tex]m>16[/tex].

Solution:
We need [tex]D<0[/tex]:
[tex]D=(4k-1)^2-8(2k^2-1)=16k^2-8k+1-16k^2+8=9-8k[/tex].
[tex]9-8k<0[/tex], so [tex]k>\frac{9}{8}[/tex].
Note that the terms with [tex]k^2[/tex] cancel, so the condition is a linear inequality.

Solution:
Case [tex]a=12[/tex]: the equation becomes [tex]3=0[/tex], which has no solutions. So [tex]a=12[/tex] fits.
Case [tex]a\ne 12[/tex]: the equation is quadratic and we need [tex]\frac{D}{4}<0[/tex]:
[tex]\frac{D}{4}=(a-12)^2-3(a-12)=(a-12)(a-15)<0[/tex], so [tex]12\lt a<15[/tex], i.e. [tex]a=13[/tex] or [tex]a=14[/tex].
Answer: [tex]a\in\{12, 13, 14\}[/tex]. The value [tex]a=12[/tex] is easy to miss if we only look at the discriminant.

Solution:
Let the border be [tex]x[/tex] in wide. The photo together with the border is a rectangle [tex](12+2x)[/tex] by [tex](18+2x)[/tex].
The photo has area [tex]12\times 18=216[/tex], so photo plus border has area [tex]2\times 216=432[/tex]:
[tex](12+2x)(18+2x)=432[/tex], [tex]4x^2+60x+216=432[/tex], [tex]x^2+15x-54=0[/tex].
[tex]D=225+216=441[/tex], [tex]x_{1,2}=\frac{-15\pm 21}{2}[/tex], [tex]x_1=3[/tex], [tex]x_2=-18[/tex] (rejected).
The border is [tex]3[/tex] in wide.

Solution:
Let [tex]x[/tex] skiers be expected. The planned portion was [tex]\frac{150}{x}[/tex], the actual one [tex]\frac{150}{x+10}[/tex]:
[tex]\frac{150}{x}-\frac{150}{x+10}=\frac{1}{6}[/tex].
The left side equals [tex]\frac{1500}{x(x+10)}[/tex], so [tex]x(x+10)=9000[/tex], [tex]x^2+10x-9000=0[/tex].
[tex]D=100+36000=36100=190^2[/tex], [tex]x_{1,2}=\frac{-10\pm 190}{2}[/tex], [tex]x_1=90[/tex], [tex]x_2=-100[/tex] (rejected).
[tex]90[/tex] skiers were expected ([tex]100[/tex] came).

Solution:
Let the slower crew pave [tex]x[/tex] miles per day; the faster one paves [tex]4.5-x[/tex]. The slower crew needs one day more:
[tex]\frac{10}{x}-\frac{10}{4.5-x}=1[/tex].
Multiply by [tex]x(4.5-x)[/tex]: [tex]45-10x-10x=4.5x-x^2[/tex], i.e. [tex]x^2-24.5x+45=0[/tex], or [tex]2x^2-49x+90=0[/tex].
[tex]D=2401-720=1681=41^2[/tex], [tex]x_{1,2}=\frac{49\pm 41}{4}[/tex], [tex]x_1=2[/tex], [tex]x_2=22.5[/tex].
[tex]22.5[/tex] is impossible (more than both crews together), so the crews paved [tex]2[/tex] and [tex]2.5[/tex] miles per day ([tex]5[/tex] days and [tex]4[/tex] days).

Solution:
Number of teams: each of [tex]n[/tex] teams plays [tex]n-1[/tex] home matches, so there are [tex]n(n-1)[/tex] matches.
[tex]n(n-1)=380[/tex], [tex]n^2-n-380=0[/tex], [tex]D=1+1520=1521=39^2[/tex], [tex]n=\frac{1+39}{2}=20[/tex].
Points: a match with a winner gives [tex]3[/tex] points in total, a draw gives [tex]2[/tex] points in total. Let [tex]W[/tex] matches end with a win and [tex]380-W[/tex] in a draw:
[tex]3W+2(380-W)=994[/tex], [tex]W+760=994[/tex], [tex]W=234[/tex].
Answer: [tex]20[/tex] teams, [tex]234[/tex] matches ended with a win.

Solution:
Let there be [tex]x[/tex] rows with [tex]y[/tex] seats each before the renovation: [tex]xy=320[/tex] and [tex](x+1)(y+4)=420[/tex].
Expanding the second equation: [tex]xy+4x+y+4=420[/tex], so [tex]4x+y=96[/tex] and [tex]y=96-4x[/tex].
[tex]x(96-4x)=320[/tex], [tex]x^2-24x+80=0[/tex], [tex]x_1=20[/tex], [tex]x_2=4[/tex].
For [tex]x=4[/tex] a row would have [tex]y=80>25[/tex] seats - impossible. So [tex]x=20[/tex], [tex]y=16[/tex].
After the renovation there are [tex]21[/tex] rows of [tex]20[/tex] seats.

Solution:
Let the units digit be [tex]a[/tex]; the tens digit is [tex]a-2[/tex]. The number is [tex]10(a-2)+a=11a-20[/tex] and the digit sum is [tex]2a-2[/tex].
[tex](11a-20)(2a-2)=952[/tex], [tex]22a^2-62a+40=952[/tex], [tex]11a^2-31a-456=0[/tex].
[tex]D=961+20064=21025=145^2[/tex], [tex]a_{1,2}=\frac{31\pm 145}{22}[/tex], [tex]a_1=8[/tex], the other root is negative.
The number is [tex]68[/tex]. Check: [tex]68\times 14=952[/tex].

Solution:
Let the sheet be [tex]L[/tex] by [tex]W[/tex]; then [tex]L+W=48[/tex]. The base of the box is [tex](L-8)[/tex] by [tex](W-8)[/tex] and its height is [tex]4[/tex]:
[tex]4(L-8)(W-8)=768[/tex], so [tex](L-8)(W-8)=192[/tex].
With [tex]W=48-L[/tex]: [tex](L-8)(40-L)=192[/tex], [tex]L^2-48L+512=0[/tex], [tex]D=2304-2048=256[/tex], [tex]L=32[/tex] or [tex]L=16[/tex].
The sheet is [tex]32[/tex] by [tex]16[/tex], so the base of the box is [tex]24[/tex] by [tex]8[/tex] and the ratio is [tex]24:8=3[/tex].

Solution:
Let [tex]q=1+\frac{p}{100}[/tex], where [tex]p\%[/tex] is the rate. After a year the deposit grows [tex]q[/tex] times:
[tex](1640q-882)q=882[/tex], so [tex]1640q^2-882q-882=0[/tex], or [tex]820q^2-441q-441=0[/tex].
[tex]D=441^2+4\times 820\times 441=441\times 3721=1281^2[/tex], [tex]q=\frac{441+1281}{1640}=\frac{1722}{1640}=1.05[/tex] (the other root is negative).
So the interest rate is [tex]5\%[/tex].

Solution:
Let [tex]f(x)=ax^2+bx+c[/tex]. Then [tex]f(0)=c<0[/tex] and [tex]f(1)=a+b+c>0[/tex].
The graph of [tex]f[/tex] is an unbroken curve (a parabola); it is below the [tex]x[/tex]-axis at [tex]x=0[/tex] and above it at [tex]x=1[/tex], so it crosses the axis between [tex]0[/tex] and [tex]1[/tex]. Hence the equation has a root in [tex](0, 1)[/tex].
The other statements may fail: [tex]-x^2+3x-1=0[/tex] fits the conditions with [tex]a<0[/tex]; [tex]3x^2-x-1=0[/tex] fits them with [tex]b<0[/tex]; and real roots always exist, as shown.

Solution:
Let [tex]x_0[/tex] be a common root. Subtract the second equality from the first:
[tex]-qx_0+5-5x_0+q=0[/tex], i.e. [tex](q+5)(1-x_0)=0[/tex].
If [tex]x_0=1[/tex]: [tex]1-q+5=0[/tex], so [tex]q=6[/tex]. Check: [tex]x^2-6x+5=0[/tex] has roots [tex]1[/tex], [tex]5[/tex], and [tex]x^2+5x-6=0[/tex] has roots [tex]1[/tex], [tex]-6[/tex].
If [tex]q=-5[/tex]: both equations become [tex]x^2+5x+5=0[/tex] ([tex]D=5>0[/tex]), so they have common roots.
Answer: [tex]q=6[/tex] or [tex]q=-5[/tex].

Solution:
[tex](px+r)^2=p^2x^2+2prx+r^2[/tex] has a positive leading coefficient and a zero discriminant. So we need [tex]a-1>0[/tex] and [tex]\frac{D}{4}=0[/tex].
[tex]\frac{D}{4}=a^2-(a-1)(3a-2)=-2a^2+5a-2=0[/tex], i.e. [tex]2a^2-5a+2=0[/tex], [tex]a=2[/tex] or [tex]a=\frac{1}{2}[/tex].
For both values the equation [tex]f(x)=0[/tex] has two equal roots, but:
[tex]a=2[/tex]: [tex]f(x)=x^2+4x+4=(x+2)^2[/tex] - a perfect square;
[tex]a=\frac{1}{2}[/tex]: [tex]f(x)=-\frac{1}{2}x^2+x-\frac{1}{2}=-\frac{1}{2}(x-1)^2[/tex] - negative leading coefficient, not a square.
Answer: [tex]a=2[/tex].

Solution:
Case [tex]a=0[/tex]: the equation is linear, [tex]2x+3=0[/tex], with the single root [tex]x=-\frac{3}{2}[/tex].
Case [tex]a\ne 0[/tex]: the equation is quadratic; it has one (double) root when [tex]\frac{D}{4}=0[/tex]:
[tex]\frac{D}{4}=(a+1)^2-a(a+3)=1-a=0[/tex], so [tex]a=1[/tex]. Then [tex]x^2+4x+4=0[/tex], [tex]x=-2[/tex].
Answer: [tex]a=0[/tex] or [tex]a=1[/tex].
(For [tex]a<1[/tex], [tex]a\ne 0[/tex] there are two roots, for [tex]a>1[/tex] none.)