Step-by-step explanation:
no, no, you use the equation to identify a, b and c.
then you use the formula with these values to get the solutions to the equation (the values of x for which the equation delivers 0).
always remember, a quadratic equation always has 2 solutions. sometimes at least one of them are not a real number, sometimes the 2 springs are equal, ...
x² - 4x - 2 = 0
a, b, c are the factors of the terms in x and the constant term.
so,
a = 1
b = -4
c = -2
x = (-b ± sqrt(b² - 4ac))/(2a)
x = (4 ± sqrt((-4)² - 4×1×-2))/(2×1) =
= (4 ± sqrt(16 + 8))/2 = (4 ± sqrt(24))/2 =
= (4 ± sqrt(4×6))/2 = (4 ± 2×sqrt(6)) / 2 =
= 2 ± sqrt(6)
x1 = 2 + sqrt(6)
x2 = 2 - sqrt(6)
I think it’s .22 I’m not sure hope this helps :)
Answer:
a) 2x² - x + 3 = x(2x + 1) - 2x
<=> 2x² - x + 3 = 2x² + x - 2x
<=> 3 - x = - x
<=> 3 = 0
=> no solution
b) x/2 - x/3 - x/4 = 1/12
=> the euqation has the solution x = 1
c) |x - 5| = 2|x|
=> x - 5 = 2x
or x - 5 = -2x
<=> x = -5
or 3x = 5
<=> x = -5
x = -5or x = 5/3
d) defined conditions: 2 - x 》 0
<=> x 《 2
we have:
<=> x = 0 (because x 《 2)
e)
<=> x = -1/4
The answer is 0.02 because the 2 is in the hundredths place
