Jika pada keadaan dasar elektron terakhir dari suatu atom adalah: n = 4 ; l = 2 ; m=0 ; s = -1/2 Maka jumlah elektron tidak berpasangan pada atom tersebut adalah

7 x 2 − 144 = − x 4 x 4 + 7 x 2 − 144 = 0 ( x 2 ) 2 + 7 x 2 − 144 = 0 s u b s t i t u t e : t = x 2 ≥ 0 t 2 + 7 t − 144 = 0 Δ = 7 2 − 4 ⋅ 1 ⋅ ( − 144 ) = 49 + 576 = 625 ; Δ ​ = 625 ​ = 25 t 1 ​ = 2 ⋅ 1 − 7 − 25 ​ = 2 − 32 ​ = − 16 < 0 ; t 2 ​ = 2 ⋅ 1 − 7 + 25 ​ = 2 18 ​ = 9 x 2 = 9 ⟺ x = ± 9 ​ → x = − 3 ∨ x = 3

Answered by Anonymous | 2024-06-10

Add x^4 to the left side and factor. x^4 + 7x^2 - 144 = 0 then factor (x^2 + 16)(x^2 - 9) = 0 set both factors = 0 x^2 + 16 = 0 x^2 = -16 square root each side x = + and - 4i x^2 - 9 = 0 x^2 = 9 square root each side x = + and - 3

Answered by casecarrie | 2024-06-10

The zeros of the equation 7 x 2 − 144 = − x 4 are x = − 3 and x = 3 . These solutions were found by rearranging the equation and using the quadratic formula. The calculation revealed that only t = 9 gives valid real solutions for x .
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Answered by Anonymous | 2024-12-24

Jawaban:Ada di penjelasan.Penjelasan:Berdasarkan soal,n = 4; l = 2; m = 0; s = -1/2 adalah konfigurasi elektron valensinya 4p⁵.Jadi elektron yang tidak berpasangan ada 1.

Answered by jouharidimas | 2025-07-04