How do you solve $4 x^{2} + 8 x + 4 = 0$ $4x^2 + 8x + 4 = 0$ ?

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Answer 1 · 2018-03-14T02:20:22

The solution is $x = - 1$ $x=-1$ .

First, divide both sides by $4$ $4$ . Then, factor. Lastly, square root both sides and solve for $x$ $x$ . Here's what that looks like:

$4 x^{2} + 8 x + 4 = 0$ $4x^2+8x+4=0$

$\frac{4 x^{2} + 8 x + 4}{4} = \frac{0}{4}$ $(4x^2+8x+4)/4=0/4$

$x^{2} + 2 x + 1 = 0$ $x^2+2x+1=0$

$x^{2} + x + x + 1 = 0$ $x^2+x+x+1=0$

$x \cdot x + x \cdot 1 + 1 \cdot x + 1 \cdot 1 = 0$ $color(red)x*x+color(red)x*1+color(blue)1*x+color(blue)1*1=0$

$x (x + 1) + 1 (x + 1) = 0$ $color(red)x(x+1)+color(blue)1(x+1)=0$

$(x + 1) (x + 1) = 0$ $(color(red)x+color(blue)1)(x+1)=0$

${(x + 1)}^{2} = 0$ $(x+1)^2=0$

$\sqrt{{(x + 1)}^{2}} = \sqrt{0}$ $sqrt((x+1)^2)=sqrt0$

$x + 1 = 0$ $x+1=0$

$x = - 1$ $x=-1$

This is the answer. Hope this helped!

How do you solve 4x2+8x+4=04x^2 + 8x + 4 = 0?