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Soal matematika persamaan kuadrat kelas X ?

Jika a dan b akar akar persamaan kuadrat 2x^2 + x - 4 = 0 maka nilai dari

(4a^2 + 2a + 1)(6b^2 + 3b -1)...

a. 72 b. 76 c. 84 d. 96 e. 99

Jika a dan b akar akar persamaan kuadrat x^2 - 12x + 4 = 0, maka nilai dari

(√a + √b) =

a. 2 b. 3 c. 4 n. 2√2 e. 2√3

apabila m akar persamaan kuadrat x^2 + 3x - 1 = 0 maka nilai dari m^3 - 10m =

a. -6 b. -4 c. -3 d. -2 e. 4

ane bakal rate 5, terimakasih

1 Answer

Rating
  • 7 years ago
    Favorite Answer

    Jawab

    1. Akar akar dari persamaan kuadrat 2x² + x - 4 = 0, adalah

    a= 1/4 (-1-√33)........... b= 1/4 (√(33)-1)

    jadi nilai (4a² + 2a + 1)(6b² + 3b -1) =

    (4[1/4 (-1-√33)]² + 2[ 1/4 (-1-√33)]+ 1) • (6[1/4 (√(33)-1)]² + 3[1/4 (√(33)-1)] -1) = .. 99

    jawaban nya : e.99

    ---------------------

    2. Akar akar dari persamaan kuadrat x² - 12x + 4 = 0, adalah

    a = 6-4 √2 ..... b = 6+4 √2

    jadi nilai (√a + √b) =

    √(6 - 4√2) + √(6 + 4√2) = .....4

    jadi jawaban nya : c.4

    ----------------------------

    3. akar persamaan kuadrat x² + 3x - 1 = 0

    m1 = 1/2 (-3-√13)........................m2 = 1/2 (√13-3)

    jadi nilai m³ - 10m = .....[kita ambil m1 sebagai contoh]

    m³ - 10m =

    [1/2 (-3-√13)]³ - 10[1/2 (-3-√13)]

    = - 3

    jadi jawaban nya : c. -3

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