Quadratic Equations: Difficult Problems with Solutions

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^2-9x+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:
[tex]x^2-4x+4+2x=7x-14[/tex]
[tex]x^2-9x+18=0[/tex]. The discriminant is [tex]D=9^2-4.18=9(9-4.2)=9.1=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:
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:
x 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]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:
[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]154 +11x +14x +x^2 = 304[/tex] [tex]x^2 +25x +154=304[/tex] [tex]x^2+25x+154-304=0[/tex] [tex]x^2+25x-150=0[/tex] [tex]x^2+30x-5x-150=0[/tex] [tex]x(x+30) - 5(x+30)=0[/tex] [tex](x-5) (x+30) = 0[/tex]
Therfore, [tex]x-5=0 \Leftrightarrow x=5[/tex]
[tex]x+30=0 \Leftrightarrow x=-30[/tex]