Question

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

Answer 1

`x=-1+-1/2sqrt3`

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"`

Answer 2

`x = -1 + 1/2sqrt3, -1 - 1/2sqrt3`

Explanation:

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

`ax^2 + bx + c = 0` (quadratic form)
`(4)x^2 + (8)x + (1) = 0` (your equation)

So:
`a = 4`
`b = 8`
`c = 1`

Now plug those in to the quadratic formula and solve:
`x = (-b \pm sqrt(b^2-4ac)) / (2a)`

`x = (-(8) \pm sqrt(8^2-4(4)(1))) / (2(4)`

`x = (-8 \pm sqrt(64-(16))) / (8)`

`x = (-8 \pm sqrt(48)) / 8`

Since `48` isn't an even square root, we will factor:

`x = (-8 \pm sqrt(48)) / 8 rarr x = (-8 \pm sqrt(3 * 4^2)) / 8 rarr x = (-8 \pm 4sqrt(3)) / 8`

Now divide by `4` to simplify:

`(cancel"-8"color(red)2 \pm cancel4color(red)1sqrt(3)) / (cancel8 color(red)2) rarr (-2 \pm 1sqrt(3)) / 2 rarr -1 \pm 1/2sqrt3`

`x = -1 + 1/2sqrt3, -1 - 1/2sqrt3`

Answer 3

(-2±√3)/2

Explanation:

We will solve this quadratic equation using quadratic formula, which goes as follows :-
4`x`²+8`x`+1=0
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
`(-b± √D) /2a`
Now let the two solutions of this equation be A and B
Therefore,
A=`(-b-√D) /2a` and B=`(-b+√D) /2a`
A=`(-8-√48)/2×4` and B=`(-8+√48) /2×4`
A=`(-8-√48)/8` and B=`(-8+√48)/8`
A=`(-8-4√3)/8` and B=`(-8+4√3)/8`
A=`(-2-√3)/2` and B=`(-2+√3)/2`