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=