What are the solutions to the equation 2k^2+3k^2+4k^2=116 ?

The solutions to the equation 2k^2+3k^2+4k^2=116 are k=(2sqrt(29))/3 ,k=-(2sqrt(29))/3

Find the zeros of 2k^2+3k^2+4k^2=116

The zeros of 2k^2+3k^2+4k^2=116 are k=(2sqrt(29))/3 ,k=-(2sqrt(29))/3

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

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

Decimal Notation

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

Solution steps

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

Add similar elements:

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

9 k^{2} = 116

Divide both sides by

9

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

For

x^{2} = f (a)

the solutions are

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}

3 q^{2} - 112 q + 930 = 0

11.48 = - 0.5 x^{2} + 0.5 x + 15

x (18 - 2 x) = 40

5 x^{2} = 4 x - 5

5 x^{2} = 4 x + 3

What are the solutions to the equation 2k^2+3k^2+4k^2=116 ?
The solutions to the equation 2k^2+3k^2+4k^2=116 are k=(2sqrt(29))/3 ,k=-(2sqrt(29))/3
Find the zeros of 2k^2+3k^2+4k^2=116
The zeros of 2k^2+3k^2+4k^2=116 are k=(2sqrt(29))/3 ,k=-(2sqrt(29))/3