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 :-
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#