# How do you solve #0= 4x ^ { 2} + 8x + 1#?

##### 3 Answers

#### Explanation:

#"given a quadratic equation in "color(blue)"standard form"#

#•color(white)(x)ax^2+bx+c=0color(white)(x);a!=0#

#"we can solve for x using the "color(blue)"quadratic formula"#

#•color(white)(x)x=(-b+-sqrt(b^2-4ac))/(2a)#

#0=4x^2+8x+1" is in standard form"#

#"with "a=4,b=8" and "c=1#

#rArrx=(-8+-sqrt(64-16))/8#

#color(white)(rArrx)=(-8+-sqrt48)/8=(-8+-4sqrt3)/8#

#rArrx=-1+-1/2sqrt3larrcolor(red)"exact solutions"#

#### Explanation:

We will solve using the quadratic formula, since your equation is in quadratic form:

So:

Now plug those in to the quadratic formula and solve:

Since

Now divide by

(-2±√3)/2

#### Explanation:

We will solve this quadratic equation using quadratic formula, which goes as follows :-

4

In this equation

a=4

b=8

c=1

First we will find the discriminant(D) for this equation

D=b²-4ac

D= 8²-4×4×1

D=64-16

D=48

The solution of this equation according to quadratic formula is given as

Now let the two solutions of this equation be A and B

Therefore,

A=

A=

A=

A=

A=