Diberikan persamaan [tex] x^2+\text{a}x+8 =0[/tex].
Maka, selisih akar-akarnya:
[tex]\quad \begin{aligned} x_1-x_2 &= \frac{\sqrt{b^2-4ac} }{a } \\ &= \frac{ \sqrt {(2\rm a)^2-4(1)(2)}}{ 1}\\ & =\sqrt{4\rm a^2-8} \end{aligned} [/tex]
Substitusikan hasilnya ke ekspresi di soal:
[tex]\begin{align} (x_1-x_2)^2 &= 4\rm a \\ (\sqrt{4\rm a^2-8})^2 &= 4\rm a \\ \rm4a^2-8 &= 4\rm a \\ \rm 4a^2-4a-8 &= 0 \\\rm a^2-a-2 &= 0 \\ \rm(a+1)(a-2) &= 0 \\ \quad\rm a=-1 \text{ atau } a&= 2\end{aligned} [/tex]
Untuk [tex]\rm a>0, [/tex] maka [tex]\rm a=2. [/tex] (Opsi D)
[tex] \Large\colorbox{Orange}{\color{white}{\#ImaginaryMe}} [/tex]
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Diberikan persamaan [tex] x^2+\text{a}x+8 =0[/tex].
Maka, selisih akar-akarnya:
[tex]\quad \begin{aligned} x_1-x_2 &= \frac{\sqrt{b^2-4ac} }{a } \\ &= \frac{ \sqrt {(2\rm a)^2-4(1)(2)}}{ 1}\\ & =\sqrt{4\rm a^2-8} \end{aligned} [/tex]
Substitusikan hasilnya ke ekspresi di soal:
[tex]\begin{align} (x_1-x_2)^2 &= 4\rm a \\ (\sqrt{4\rm a^2-8})^2 &= 4\rm a \\ \rm4a^2-8 &= 4\rm a \\ \rm 4a^2-4a-8 &= 0 \\\rm a^2-a-2 &= 0 \\ \rm(a+1)(a-2) &= 0 \\ \quad\rm a=-1 \text{ atau } a&= 2\end{aligned} [/tex]
Untuk [tex]\rm a>0, [/tex] maka [tex]\rm a=2. [/tex] (Opsi D)
[tex] \Large\colorbox{Orange}{\color{white}{\#ImaginaryMe}} [/tex]