Quadratic Equations: Problems with Solutions

Solution:
D = 3² - 4 ⋅ 4 = -7 < 0
So the equation has 0 real roots.

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:
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:
[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.