Home > Algebra > Comparing Methods for Solving Quadratics

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

Answer 1

The solution is #x=-1#.

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

#4x^2+8x+4=0#

#(4x^2+8x+4)/4=0/4#

#x^2+2x+1=0#

#x^2+x+x+1=0#

#color(red)x*x+color(red)x*1+color(blue)1*x+color(blue)1*1=0#

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

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

#(x+1)^2=0#

#sqrt((x+1)^2)=sqrt0#

#x+1=0#

#x=-1#

I hope this helped. Here's the solution!

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Hazel Anglin

To solve the quadratic equation (4x^2 + 8x + 4 = 0), you can use the quadratic formula:

[x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}]

Where (a = 4), (b = 8), and (c = 4). Substituting these values into the formula:

[x = \frac{{-(8) \pm \sqrt{{(8)^2 - 4(4)(4)}}}}{{2(4)}}]

[x = \frac{{-8 \pm \sqrt{{64 - 64}}}}{{8}}]

[x = \frac{{-8 \pm \sqrt{{0}}}}{{8}}]

[x = \frac{{-8}}{{8}}]

[x = -1]

So, the solution to the quadratic equation (4x^2 + 8x + 4 = 0) is (x = -1).

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.