What are the solutions to the equation (-k^2)/4 =(k^2+5k)/2 ?

The solutions to the equation (-k^2)/4 =(k^2+5k)/2 are k=0,k=-10/3

Find the zeros of (-k^2)/4 =(k^2+5k)/2

The zeros of (-k^2)/4 =(k^2+5k)/2 are k=0,k=-10/3

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\frac{- k ^{2}}{4} = \frac{k ^{2} + 5 k}{2}

k = 0, k = - \frac{10}{3}

Decimal Notation

k = 0, k = - 3.33333 \dots

Solution steps

\frac{- k ^{2}}{4} = \frac{k ^{2} + 5 k}{2}

Find Least Common Multiplier of

4, 2 : 4

Multiply by LCM=

4

\frac{- k ^{2}}{4} \cdot 4 = \frac{k ^{2} + 5 k}{2} \cdot 4

Simplify

- k^{2} = 2 (k^{2} + 5 k)

Expand

2 (k^{2} + 5 k) : 2 k^{2} + 10 k

- k^{2} = 2 k^{2} + 10 k

Switch sides

2 k^{2} + 10 k = - k^{2}

Move

k^{2}

to the left side

3 k^{2} + 10 k = 0

Solve with the quadratic formula

k_{1, 2} = \frac{- 10 \pm 1 0 ^{2} - 4 \cdot 3 \cdot 0}{2 \cdot 3}

1 0^{2} - 4 \cdot 3 \cdot 0 = 10

k_{1, 2} = \frac{- 10 \pm 10}{2 \cdot 3}

Separate the solutions

k_{1} = \frac{- 10 + 10}{2 \cdot 3}, k_{2} = \frac{- 10 - 10}{2 \cdot 3}

k = \frac{- 10 + 10}{2 \cdot 3} : 0

k = \frac{- 10 - 10}{2 \cdot 3} : - \frac{10}{3}

The solutions to the quadratic equation are:

k = 0, k = - \frac{10}{3}

12 x - 3 \leq - 4 x + 5

(w + 7)^{2} = 2 w^{2} + 7 w + 55

\frac{n ^{2} - 13 n + 42}{4 n - 28}

r^{2} + r + 1 = 0

z^{2} - z - 30

What are the solutions to the equation (-k^2)/4 =(k^2+5k)/2 ?
The solutions to the equation (-k^2)/4 =(k^2+5k)/2 are k=0,k=-10/3
Find the zeros of (-k^2)/4 =(k^2+5k)/2
The zeros of (-k^2)/4 =(k^2+5k)/2 are k=0,k=-10/3