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

3 Answers
Mar 20, 2018

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"

Mar 20, 2018

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

Mar 20, 2018

(-2±√3)/2

Explanation:

We will solve this quadratic equation using quadratic formula, which goes as follows :-
4x²+8x+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