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

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

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

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

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

k = 4, k = - 5

Solution steps

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

k - 4 = 0 or k + 5 = 0

Solve

k - 4 = 0 : k = 4

Solve

k + 5 = 0 : k = - 5

The solutions to the quadratic equation are:

k = 4, k = - 5

(\frac{3 x}{2})^{4}

3 x^{2} - 114 = - x

x = 1^{2}

5 x^{2} + 9 x y + 4 y^{2}

\frac{1}{2} \cdot 4 \cdot 5

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