Mohon bantuannya minna-san ! >

Ia. [tex] x^2+3x-18 = 0 [/tex]
[tex] (x+6)(x-3) = 0 [/tex]
[tex] x = -6 \vee x=3 [/tex]

Ib. [tex] x^2 - 8x + 15 = 0 [/tex]
[tex] x^2 - 8x + 16 = 1 [/tex]
[tex] (x-4)^2 = 1 [/tex]
[tex] x-4 = \pm \sqrt{1} [/tex]
[tex] x = 4 \pm \sqrt{1} [/tex]

Ic. [tex] 2x^2+4x-1=0 [/tex]
[tex] x_{1,2} = \frac{-4\pm \sqrt{24}}{4} [/tex]

IIa.[tex] x^2 - 3 = 0 [/tex]
Difference of two square, therefore 2 solution on Real plane.

IIb.[tex] 2x-8x+8 =0 [/tex]
[tex] D = 64 - 64 = 0 [/tex]
Only one solution on Real plane.

IIc.[tex] 6x^2+2x+1=0 [/tex]
[tex] D = 4 - 32 = -28 [/tex]
No solution on Real plane, both solution exist on complex plane.

III[tex]3x^2-px+3=0 [/tex]
[tex](\sqrt{3}x-\sqrt{3})(\sqrt{3}x-\sqrt{3}) = 0 [/tex]
[tex] (a-b)(a-b) = a^2 - 2ab + b^2 [/tex]
[tex] -px = -2\sqrt{3}x \sqrt{3} [/tex]
[tex] p=6 [/tex]

IV [tex] 2x^2+4x-3=0 [/tex]
[tex] x_{1,2} = \frac{-4 \pm 2\sqrt{10}}{4} [/tex]
[tex] x_1 = -1 +\frac{\sqrt{10}}{2} \vee x_2 = -1 - \frac{\sqrt{10}}{2} [/tex]
*Note [tex] a^2-b^2=(a+b)(a-b) [/tex]

[tex] (x_1+x_2)^2 = (-2)^2 = 4 [/tex]

[tex] 2x_1 x_2 = 2(-1+\frac{\sqrt{10}}{2})(-1-\frac{\sqrt{10}}{2}) [/tex]
[tex] 2x_1 x_2 = 2(1 - \frac{10}{4}) = -3 [/tex]

[tex]\frac{1}{x_1}+\frac{1}{x_2} = \frac{ x_2 + x_1 }{x_1\cdot x_2} [/tex]
[tex]\frac{1}{x_1}+\frac{1}{x_2} = \frac{-2}{\frac{-3}{2}} [/tex]
[tex]\frac{1}{x_1}+\frac{1}{x_2} = \frac{4}{3} [/tex]

V [tex] 4x^2 -2x -3 =0 [/tex]
[tex] x_{1,2} = \frac{2 \pm 2\sqrt{13}}{8} [/tex]
[tex] x_1 = \frac{1}{4} + \frac{\sqrt{13}}{4} \vee x_2 = \frac{1}{4} - \frac{\sqrt{13}}{4} [/tex]

[tex] ( x - 4x_1)(x - 4x_2)= x^2 - 4x(\cdot x_1 +\cdot x_2) + 16 x_1 x_2 [/tex]
[tex] (x-4x_1)(x-4x_2) = x^2 -4x(\frac{1}{2})+ 16(\frac{1-13}{16}) [/tex]
[tex] (x-4x_1)(x-4x_2) = x^2 - 2x - 12 [/tex]

[tex] x'_{1,2}=\frac{5 \pm \sqrt{13}}{4} [/tex]
[tex] (x-x'_1)(x-x'_2) = x^2 - x(x'_1+x'_2) + x'_1 x'_2 [/tex]
[tex] (x-x'_1)(x-x'_2) = x^2 - x(\frac{10}{4}) + \frac{25-13}{16} [/tex]
[tex] (x-x'_1)(x-x'_2) = x^2 - \frac{10}{4}x + \frac{3}{4} [/tex]





*xddd.