Hitunglah nilai k agar persamaan 3x2+(k−2)x+k+1=0 mempunyai akar-akar berkebalikan!

Pertanyaan

Hitunglah nilai $k$ agar persamaan $3x^2+\left(k-2\right)x+k+1=0$ mempunyai akar-akar berkebalikan!

N. Puspita

Master Teacher

Jawaban terverifikasi

Jawaban

nilai

agar persamaan 3 x squared plus open parentheses k minus 2 close parentheses x plus k plus 1 equals 0

mempunyai akar berkebalikan adalah 2.

Pembahasan

Dari soal diketahui

$\begin{array}{l}3x^2+\left(k-2\right)x+k+1=0\\a=3\\b=\left(k-2\right)\\c=k+1\end{array}$

Maka

$table attributes columnalign right center left columnspacing 0px end attributes row cell x subscript 1 plus x subscript 2 end cell equals cell negative b over a end cell row blank equals cell negative fraction numerator open parentheses k minus 2 close parentheses over denominator 3 end fraction end cell row blank equals cell fraction numerator negative k plus 2 over denominator 3 end fraction end cell row blank blank blank row cell x subscript 1 times x subscript 2 end cell equals cell c over a end cell row blank equals cell fraction numerator k plus 1 over denominator 3 end fraction end cell end table$

Karena akar-akarnya saling berkebalikan, maka x subscript 1 equals 1 over x subscript 2

Sehingga diperoleh

$table attributes columnalign right center left columnspacing 0px end attributes row cell x subscript 1 times x subscript 2 end cell equals cell fraction numerator k plus 1 over denominator 3 end fraction end cell row cell 1 over x subscript 2 times x subscript 2 end cell equals cell fraction numerator k plus 1 over denominator 3 end fraction end cell row 1 equals cell fraction numerator k plus 1 over denominator 3 end fraction end cell row 3 equals cell k plus 1 end cell row k equals cell 3 minus 1 end cell row blank equals 2 end table$

Jadi, nilai agar persamaan mempunyai akar berkebalikan adalah 2.