What are the solutions to the equation 2(4x+1)(2x+1)=0 ?

The solutions to the equation 2(4x+1)(2x+1)=0 are x=-1/4 ,x=-1/2

Find the zeros of 2(4x+1)(2x+1)=0

The zeros of 2(4x+1)(2x+1)=0 are x=-1/4 ,x=-1/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2 (4 x + 1) (2 x + 1) = 0

x = - \frac{1}{4}, x = - \frac{1}{2}

Solution steps

2 (4 x + 1) (2 x + 1) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

4 x + 1 = 0 or 2 x + 1 = 0

Solve

4 x + 1 = 0 : x = - \frac{1}{4}

Solve

2 x + 1 = 0 : x = - \frac{1}{2}

The solutions to the quadratic equation are:

x = - \frac{1}{4}, x = - \frac{1}{2}

(- 2 + \frac{4 i}{n})^{2}

lo g_{7} (\frac{1}{49})

∣ 4 x - 1 ∣ \geq 0

6 + 10 k = 256

1 0^{2} + 1 0^{2} = c^{2}

What are the solutions to the equation 2(4x+1)(2x+1)=0 ?
The solutions to the equation 2(4x+1)(2x+1)=0 are x=-1/4 ,x=-1/2
Find the zeros of 2(4x+1)(2x+1)=0
The zeros of 2(4x+1)(2x+1)=0 are x=-1/4 ,x=-1/2