What are the solutions to the equation (4)^3-k^2(4)^2-8k(4)-16=0 ?

The solutions to the equation (4)^3-k^2(4)^2-8k(4)-16=0 are k=-3,k=1

Find the zeros of (4)^3-k^2(4)^2-8k(4)-16=0

The zeros of (4)^3-k^2(4)^2-8k(4)-16=0 are k=-3,k=1

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(4)^{3} - k^{2} (4)^{2} - 8 k (4) - 16 = 0

k = - 3, k = 1

Solution steps

(4)^{3} - k^{2} (4)^{2} - 8 k (4) - 16 = 0

Expand

4^{3} - k^{2} \cdot 4^{2} - 8 k (4) - 16 : - 16 k^{2} - 32 k + 48

- 16 k^{2} - 32 k + 48 = 0

Solve with the quadratic formula

k_{1, 2} = \frac{- ( - 32 ) \pm ( - 32 ) ^{2} - 4 ( - 16 ) \cdot 48}{2 ( - 16 )}

(- 32)^{2} - 4 (- 16) \cdot 48 = 64

k_{1, 2} = \frac{- ( - 32 ) \pm 64}{2 ( - 16 )}

Separate the solutions

k_{1} = \frac{- ( - 32 ) + 64}{2 ( - 16 )}, k_{2} = \frac{- ( - 32 ) - 64}{2 ( - 16 )}

k = \frac{- ( - 32 ) + 64}{2 ( - 16 )} : - 3

k = \frac{- ( - 32 ) - 64}{2 ( - 16 )} : 1

The solutions to the quadratic equation are:

k = - 3, k = 1

8 a^{2} - 10 a + 2 = 0

16 a^{2} - 25 = 0

x, p x^{2} - p (p + 3) x + 2 p (p + 1) = 0

x, y = x^{2} + 4 x + 9

x, h = - \frac{x}{2} (x - r) + 2

What are the solutions to the equation (4)^3-k^2(4)^2-8k(4)-16=0 ?
The solutions to the equation (4)^3-k^2(4)^2-8k(4)-16=0 are k=-3,k=1
Find the zeros of (4)^3-k^2(4)^2-8k(4)-16=0
The zeros of (4)^3-k^2(4)^2-8k(4)-16=0 are k=-3,k=1