2k^2+3k^2+4k^2=116

解

2 k^{2} + 3 k^{2} + 4 k^{2} = 116

k = \frac{2 29}{3}, k = - \frac{2 29}{3}

十進法表記

k = 3.59010 \dots, k = - 3.59010 \dots

解答ステップ

2 k^{2} + 3 k^{2} + 4 k^{2} = 116

類似した元を足す:

2 k^{2} + 3 k^{2} + 4 k^{2} = 9 k^{2}

9 k^{2} = 116

以下で両辺を割る

9

k^{2} = \frac{116}{9}

x^{2} = f (a)

の場合, 解は

x = f (a), - f (a)

k = \frac{116}{9}, k = - \frac{116}{9}

\frac{116}{9} = \frac{2 29}{3}

- \frac{116}{9} = - \frac{2 29}{3}

k = \frac{2 29}{3}, k = - \frac{2 29}{3}