Quadratic Equations: Problems with Solutions

Solution:
A number is a root if it turns the equation into a true equality. Substitute each number:
[tex]x=-5[/tex]: [tex](-5)^2+2\times(-5)=25-10=15[/tex] - true.
[tex]x=1[/tex]: [tex]1+2=3\ne 15[/tex] - false.
[tex]x=-0.5[/tex]: [tex]0.25-1=-0.75\ne 15[/tex] - false.
[tex]x=3[/tex]: [tex]9+6=15[/tex] - true.
The roots are [tex]-5[/tex] and [tex]3[/tex].

Solution:
This is an incomplete quadratic equation ([tex]b=0[/tex]), so we do not need the discriminant.
[tex]25x^2=16[/tex], [tex]x^2=\frac{16}{25}[/tex], so [tex]x=\pm\frac{4}{5}[/tex].
Another way: [tex](5x-4)(5x+4)=0[/tex].

Solution:
Factor out [tex]x[/tex]: [tex]x(5x-1)=0[/tex].
A product is zero when one of the factors is zero: [tex]x=0[/tex] or [tex]5x-1=0[/tex], i.e. [tex]x=\frac{1}{5}[/tex].
Note: dividing both sides by [tex]x[/tex] would lose the root [tex]x=0[/tex].

Solution:
[tex]x^2-5x+4=x^2-4x-x+4=x(x-4)-(x-4)=(x-1)(x-4)[/tex]
so the roots of the equation are [tex]x_1=1[/tex] and [tex]x_2=4[/tex]. The greater one is obviously 4.

Solution:
[tex]x^2-3x+2=x^2-2x-x+2=x(x-2)-(x-2)=(x-1)(x-2)[/tex]
so the solutions to the equation are [tex]x_1=1, x_2=2[/tex]. The lesser one is obviously 1.

Solution:
[tex]x^2-13x+12=x^2-12x-x+12=x(x-12)-(x-12)=(x-1)(x-12)[/tex], so the roots are [tex]x_1=1[/tex] and [tex]x_2=12[/tex].

Solution:
[tex]x^2-7x+12=x^2-3x-4x+12=x(x-3)-4(x-3)=(x-3)(x-4)[/tex], so the roots are [tex]x_1=3[/tex] and [tex]x_2=4[/tex].

Solution:
[tex]x^2-12x+35=x^2-7x-5x+35=x(x-7)-5(x-7)=(x-5)(x-7)[/tex]. Its roots are 5 and 7 and the lesser one is 5.

Solution:
The discriminant is [tex]D=15^2-4\cdot 26=225-104=121=11^2[/tex]. The solutions are
[tex]x_{1,2}=\frac{15 \pm 11}{2}[/tex]

[tex]x_1=\frac{15+11}{2}=\frac{26}{2}=13[/tex]
[tex]x_2=\frac{15-11}{2}=\frac{4}{2}=2[/tex]

Solution:
The discriminant is [tex]D=14^2-4 \cdot 45=196-180=16=4^2[/tex]. The roots are [tex]x_{1,2}=\frac{-14 \pm 4}{2}[/tex].
Therefore, [tex]x_1=\frac{-14+4}{2}=\frac{-10}{2}=-5[/tex].
[tex]x_2 = \frac{-14-4}{2}=\frac{-18}{2}=-9[/tex]

Solution:
The discriminant is [tex]3^2+4\cdot 70=289=17^2[/tex]. The roots are [tex]x_{1,2}=\frac{-3 \pm 17}{2}[/tex]. Therefore [tex]x_1=\frac{-3+17}{2}=\frac{14}{2}=7[/tex] and [tex]x_2=\frac{-3-17}{2}=\frac{-20}{2}=-10[/tex]

Solution:
The left side is a perfect square: [tex]4x^2-12x+9=(2x)^2-2\times 2x\times 3+3^2=(2x-3)^2[/tex].
So [tex](2x-3)^2=0[/tex] and [tex]2x-3=0[/tex], i.e. [tex]x=\frac{3}{2}[/tex] (a double root).
Check with the discriminant: [tex]D=144-144=0[/tex].

Solution:
[tex]D = 3^{2} - 4 \times 4 = -7[/tex] < 0
So the equation has 0 real roots.

Solution:
An equation [tex]az^2+bz+c=0[/tex] has two distinct real roots exactly when [tex]D=b^2-4ac>0[/tex].
[tex]2z^2-z+1=0[/tex]: [tex]D=1-8=-7<0[/tex], no real roots.
[tex]4z^2+20z+25=0[/tex]: [tex]D=400-400=0[/tex], one double root.
[tex]3z^2+10z+7=0[/tex]: [tex]D=100-84=16>0[/tex], two distinct roots ([tex]-1[/tex] and [tex]-\frac{7}{3}[/tex]).
[tex]z^2-8z+20=0[/tex]: [tex]D=64-80=-16<0[/tex], no real roots.
Answer: [tex]3z^2+10z+7=0[/tex].

Solution:
A square of a real number is never negative, so [tex](x-1)^2+(x+3)^2\ge 0[/tex] for every [tex]x[/tex].
The right side is [tex]-10<0[/tex], so the equality is impossible: the equation has no real roots.
The same result by calculation: [tex]2x^2+4x+10=-10[/tex], [tex]x^2+2x+10=0[/tex], [tex]D=4-40<0[/tex].

Solution:
If [tex]x_1[/tex] and [tex]x_2[/tex] are the roots of [tex]ax^2+bx+c=0[/tex], then [tex]ax^2+bx+c=a(x-x_1)(x-x_2)[/tex].
[tex]3x^2-5x+2=0[/tex]: [tex]D=25-24=1[/tex], [tex]x_{1,2}=\frac{5\pm 1}{6}[/tex], so [tex]x_1=1[/tex], [tex]x_2=\frac{2}{3}[/tex].
[tex]3x^2-5x+2=3(x-1)\left(x-\frac{2}{3}\right)=(x-1)(3x-2)[/tex].
Check: [tex](x-1)(3x-2)=3x^2-2x-3x+2=3x^2-5x+2[/tex].

Solution:
Move everything to the left side and factor out [tex]x-2[/tex]:
[tex](x-2)^2-(x-2)(x-3)=0[/tex]
[tex](x-2)\left[(x-2)-(x-3)\right]=0[/tex]
[tex](x-2)\times 1=0[/tex], so [tex]x=2[/tex].
Warning: if we divide both sides by [tex]x-2[/tex], we get [tex]x-2=x-3[/tex], which has no solution, and we lose the root [tex]x=2[/tex].
Answer: [tex]x=2[/tex].

Solution:
Use the difference of squares [tex]A^2-B^2=(A-B)(A+B)[/tex]:
[tex](2x+5)^2-(x-3)^2=(2x+5-x+3)(2x+5+x-3)=(x+8)(3x+2)=3x^2+26x+16[/tex].
The equation becomes [tex]3x^2+26x+16=16[/tex], i.e. [tex]3x^2+26x=0[/tex], [tex]x(3x+26)=0[/tex].
Roots: [tex]x_1=0[/tex], [tex]x_2=-\frac{26}{3}[/tex].

Solution:
Expand: [tex](x+1)^3=x^3+3x^2+3x+1[/tex] and [tex]x^2(x+2)=x^3+2x^2[/tex].
The cubes cancel: [tex]x^3+3x^2+3x+1-x^3-2x^2=x^2+3x+1[/tex].
[tex]x^2+3x+1=x[/tex], so [tex]x^2+2x+1=0[/tex], i.e. [tex](x+1)^2=0[/tex].
Answer: [tex]x=-1[/tex] (a double root).

Solution:
Group the terms:
[tex]x^2-x-\sqrt{3}x+\sqrt{3}=x(x-1)-\sqrt{3}(x-1)=(x-1)(x-\sqrt{3})=0[/tex].
Roots: [tex]x_1=1[/tex], [tex]x_2=\sqrt{3}[/tex].
With the discriminant: [tex]D=(1+\sqrt{3})^2-4\sqrt{3}=4-2\sqrt{3}=(\sqrt{3}-1)^2[/tex], [tex]x_{1,2}=\frac{1+\sqrt{3}\pm(\sqrt{3}-1)}{2}[/tex] - the same roots.

Solution:
A double root is possible only for a quadratic equation, so [tex]a\ne 0[/tex] (for [tex]a=0[/tex] the equation is linear).
A quadratic equation has a double root when [tex]D=0[/tex]: [tex]D=16-4a=0[/tex], so [tex]a=4[/tex].
Check: [tex]4x^2+4x+1=(2x+1)^2=0[/tex], double root [tex]x=-\frac{1}{2}[/tex].

Solution:
Both squares are non-negative, and their sum is [tex]0[/tex] only when each of them is [tex]0[/tex]:
[tex]y-1=0[/tex], so [tex]y=1[/tex];
[tex]x-2y=0[/tex], so [tex]x=2y=2[/tex].
The only solution is [tex]x=2[/tex], [tex]y=1[/tex].

Solution:
Let the side of the square be [tex]x[/tex] inches. The remaining part is a rectangle with sides [tex]x[/tex] and [tex]x-3[/tex].
[tex]x(x-3)=70[/tex], so [tex]x^2-3x-70=0[/tex].
[tex]D=9+280=289[/tex], [tex]x_{1,2}=\frac{3\pm 17}{2}[/tex], [tex]x_1=10[/tex], [tex]x_2=-7[/tex].
A length cannot be negative, so the side of the square is [tex]10[/tex] inches.

Solution:
Let there be [tex]x[/tex] students in a row. Then there are [tex]x+6[/tex] rows and [tex]x(x+6)=247[/tex].
[tex]x^2+6x-247=0[/tex], [tex]D=36+988=1024=32^2[/tex], [tex]x_{1,2}=\frac{-6\pm 32}{2}[/tex].
[tex]x_1=13[/tex], [tex]x_2=-19[/tex] (rejected, the number of students is positive).
There were [tex]13[/tex] students in each row (in [tex]19[/tex] rows).

Solution:
Let there be [tex]n[/tex] players. Each of them plays [tex]n-1[/tex] games, but this way every game is counted twice (once for each player).
So the number of games is [tex]\frac{n(n-1)}{2}=45[/tex], i.e. [tex]n^2-n-90=0[/tex].
[tex]D=1+360=361[/tex], [tex]n_{1,2}=\frac{1\pm 19}{2}[/tex], [tex]n_1=10[/tex], [tex]n_2=-9[/tex] (rejected).
[tex]10[/tex] players took part.