What are the solutions to the equation (4k+5)(k+1)=0 ?

The solutions to the equation (4k+5)(k+1)=0 are k=-5/4 ,k=-1

Find the zeros of (4k+5)(k+1)=0

The zeros of (4k+5)(k+1)=0 are k=-5/4 ,k=-1

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(4 k + 5) (k + 1) = 0

k = - \frac{5}{4}, k = - 1

Solution steps

(4 k + 5) (k + 1) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

4 k + 5 = 0 or k + 1 = 0

Solve

4 k + 5 = 0 : k = - \frac{5}{4}

Solve

k + 1 = 0 : k = - 1

The solutions to the quadratic equation are:

k = - \frac{5}{4}, k = - 1

2 = ∣ x - 4 ∣

- 8 x < - 16

\frac{2 x - 1}{4} + \frac{x - 1}{5} = 2

∣ x - 6 ∣ = - 1

x^{2} - 63 = 2 x

What are the solutions to the equation (4k+5)(k+1)=0 ?
The solutions to the equation (4k+5)(k+1)=0 are k=-5/4 ,k=-1
Find the zeros of (4k+5)(k+1)=0
The zeros of (4k+5)(k+1)=0 are k=-5/4 ,k=-1