Pertanyaan

Jika x 1 dan x 2 akar-akar persamaan kuadrat x 2 − 2 x + 4 = 0 . Tentukan nilai: a. x 1 2 + x 2 2 b. ( x 1 − x 2 ) 2 c. x 1 1 + x 2 1 d. x 2 x 1 + x 1 x 2

Jika $x_{1}$ dan $x_{2}$ akar-akar persamaan kuadrat $x^{2} - 2 x + 4 = 0$ . Tentukan nilai:

a. $x_{1}^{2} + x_{2}^{2}$

b. $(x_{1} - x_{2})^{2}$

c. $\frac{1}{x _{1}} + \frac{1}{x _{2}}$

d. $\frac{x _{1}}{x _{2}} + \frac{x _{2}}{x _{1}}$ $\;\;$

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Habis dalam

Klaim

M. Claudia

Master Teacher

Mahasiswa/Alumni Universitas Nusa Cendana Kupang

Jawaban terverifikasi

Pembahasan

Ingat kembali, Sifat operasi akar persamaan kuadrat Jika dan akar-akar persamaan kuadrat Berdasarkan sifat di atas, diperoleh perhitungan sebagai berikut dan akar-akar persamaan kuadrat a. Sehingga, b. Sehingga, c. Sehingga, d. Sehingga, .

Ingat kembali,

Sifat operasi akar persamaan kuadrat

Jika x subscript 1 dan x subscript 2 akar-akar persamaan kuadrat a x squared plus b x plus c equals 0

$x subscript 1 plus x subscript 2 equals fraction numerator negative b over denominator a end fraction$

x subscript 1 space end subscript. space x subscript 2 equals c over a

Berdasarkan sifat di atas, diperoleh perhitungan sebagai berikut

x subscript 1 dan x subscript 2 akar-akar persamaan kuadrat x squared minus 2 x plus 4 equals 0

$x subscript 1 plus x subscript 2 equals fraction numerator negative left parenthesis negative 2 right parenthesis over denominator 1 end fraction equals 2$

x subscript 1 times x subscript 2 equals 4 over 1 equals 4

a. x subscript 1 squared plus x subscript 2 squared equals open parentheses x subscript 1 plus x subscript 2 close parentheses squared minus 2 x subscript 1 x subscript 2 equals open parentheses 2 close parentheses squared minus 2 times 4 equals negative 4

Sehingga, straight x subscript 1 squared plus straight x subscript 2 squared equals negative 4

b. open parentheses x subscript 1 minus x subscript 2 close parentheses squared equals left parenthesis x subscript 1 squared plus x subscript 2 squared right parenthesis minus 2 x subscript 1 x subscript 2 equals negative 4 minus 2 times 4 equals negative 12

Sehingga, left parenthesis straight x subscript 1 minus straight x subscript 2 right parenthesis squared equals negative 12

c. $1 over x subscript 1 plus 1 over x subscript 2 equals fraction numerator x subscript 2 plus x subscript 1 over denominator x subscript 1 x subscript 2 end fraction equals fraction numerator x subscript 1 plus x subscript 2 over denominator x subscript 1 x subscript 2 end fraction equals 2 over 4 equals 1 half$

Sehingga, 1 over straight x subscript 1 plus 1 over straight x subscript 2 equals 1 half

d. $x subscript 1 over x subscript 2 plus x subscript 2 over x subscript 1 equals fraction numerator x subscript 1 squared plus x subscript 2 squared over denominator x subscript 1 x subscript 2 end fraction equals fraction numerator negative 4 over denominator 4 end fraction equals negative 1$
Sehingga, straight x subscript 1 over straight x subscript 2 plus straight x subscript 2 over straight x subscript 1 equals negative 1 . space space