What are the solutions to the equation 8pi=2pir(r+3) ?

The solutions to the equation 8pi=2pir(r+3) are r=1,r=-4

Find the zeros of 8pi=2pir(r+3)

The zeros of 8pi=2pir(r+3) are r=1,r=-4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

8 π = 2 π r (r + 3)

r = 1, r = - 4

Solution steps

8 π = 2 π r (r + 3)

Move

2 π r (r + 3)

to the left side

8 π - 2 π r (r + 3) = 0

Factor

8 π - 2 π r (r + 3) : - 2 π (r - 1) (r + 4)

- 2 π (r - 1) (r + 4) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

r - 1 = 0 or r + 4 = 0

Solve

r - 1 = 0 : r = 1

Solve

r + 4 = 0 : r = - 4

The solutions to the quadratic equation are:

r = 1, r = - 4

\frac{12 k ^{2} - 54 k}{6 k} \cdot \frac{6}{81 - 18 k}

a^{2} + 3 a - 18 = 0

2 x - 3 \leq 4 x + 3 < 2 x + 5

2 X^{2} - 72 = 0

\frac{7 !}{5 ! ( 2 ! )}

What are the solutions to the equation 8pi=2pir(r+3) ?
The solutions to the equation 8pi=2pir(r+3) are r=1,r=-4
Find the zeros of 8pi=2pir(r+3)
The zeros of 8pi=2pir(r+3) are r=1,r=-4