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Persamaan  mempunyai akar-akar  dan . Dengan demikian maka  ...

Pertanyaan

Persamaan 2 to the power of 2 x end exponent minus 3 times 2 to the power of x plus 2 end exponent plus 32 equals 0 mempunyai akar-akar x subscript 1 dan x subscript 2. Dengan demikian maka x subscript 1 plus x subscript 2 equals ...

  1. 32 space

  2. 12 space

  3. space

  4. space

  5. space

Pembahasan Soal:

Diketahui:

Persamaan 2 to the power of 2 x end exponent minus 3 times 2 to the power of x plus 2 end exponent plus 32 equals 0

Ditanya:

x subscript 1 plus x subscript 2

Bentuk a p to the power of 2 x end exponent plus b p to the power of x plus c equals 0 merupakan persamaan eksponen yang dapat direduksi menjadi persamaan kuadrat.

Perlu diingat bahwa:

k to the power of open parentheses l plus m close parentheses end exponent equals k to the power of l times k to the power of m open parentheses k to the power of l close parentheses to the power of m equals k to the power of l times m end exponent 

Perhatikan perhitungan berikut:

table attributes columnalign right center left columnspacing 0px end attributes row cell 2 to the power of 2 x end exponent minus 3 times 2 to the power of x plus 2 end exponent plus 32 end cell equals 0 row cell 2 to the power of 2 x end exponent minus 3 times 2 to the power of x times 2 squared plus 32 end cell equals 0 row cell 2 to the power of 2 x end exponent minus 3 times 2 to the power of x times 4 plus 32 end cell equals 0 row cell 2 to the power of 2 x end exponent minus 12 times 2 to the power of x plus 32 end cell equals 0 row cell 2 to the power of 2 x end exponent minus 12 times 2 to the power of x plus 32 end cell equals 0 end table

Misalkan 2 to the power of x adalah a, maka:

space space space 2 to the power of 2 x end exponent minus 12 times 2 to the power of x plus 32 equals 0 open parentheses 2 to the power of x close parentheses squared minus 12 times 2 to the power of x plus 32 equals 0 space space space space space space space space a squared minus 12 a plus 32 equals 0 space space space space space space space space space open parentheses a minus 8 close parentheses open parentheses a minus 4 close parentheses equals 0 space space space space a minus 8 equals 0 space logical or space a minus 4 equals 0 space space space space space space space space space space a equals 8 space logical or space a equals 4

Karena a equals 2 to the power of x sehingga, x subscript 1 dan x subscript 2 dapat ditentukan seperti berikut:

space space a equals 8 space space space space space atau space space space space space a equals 4 2 to the power of x subscript 1 end exponent equals 8 space space space space space space space space space space space space space space 2 to the power of x subscript 2 end exponent equals 4 2 to the power of x subscript 1 end exponent equals 2 cubed space space space space space space space space space space space space space 2 to the power of x subscript 2 end exponent equals 2 squared space x subscript 1 equals 3 space space space space space space space space space space space space space space space space x subscript 2 equals 2

Hasil penjumlahan akar-akar x subscript 1 dan x subscript 2:

table attributes columnalign right center left columnspacing 0px end attributes row cell x subscript 1 plus x subscript 2 end cell equals cell 3 plus 2 end cell row blank equals 5 end table

Sehingga, hasil penjumlahan akar-akar x subscript 1 plus x subscript 2 adalah 5.

Jadi, jawaban yang tepat adalah D.

Pembahasan terverifikasi oleh Roboguru

Terakhir diupdate 19 Juli 2021

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