Starting with x1 = 2, find the third approximation x3 to the root of the equation x3 − 2x − 2 = 0.

0.8989

Step-by-step explanation:

Using the Newton's Raphson approximation formula.

Xn+1 = Xn - f(Xn)/f'(Xn)

Given f(x) = x³-2x+2

f'(x) = 3x²-2

If the initial value X1 = 2

X2 = X1 - f(X1)/f'(X1)

X2 = 2 - f(2)/f'(2)

f(2) = 2³-2(2)+2

f(2) = 8-4+2

f(2) = 6

f'(2) = 3(2)²-2

f'(2) = 10

X2 = 2- 6/10

X2 = 14/10

X2 = 1.4

X3 = X2 - f(X2)/f'(X2)

X3 = 1.4 - f(1.4)/f'(1.4)

f(1.4) = 1.4³-2(1.4)+2

f(1.4) = 2.744-2.8+2

f(1.4) = 1.944

f'(1.4) = 3(1.4)²-2

f'(1.4) = 3.880

X3 = 1.4- 1.944/3.880

X3 = 1.4 - 0.5010

X3 = 0.8989

Hence the value of X3 is 0.8989

{x = -1/2 ,   y = -1

step-by-step explanation:

solve the following system:

{2 x + y = -2

{10 x - 3 y = -2

in the first equation, look to solve for y:

{2 x + y = -2

{10 x - 3 y = -2

subtract 2 x from both sides:

{y = -2 x - 2

{10 x - 3 y = -2

substitute y = -2 x - 2 into the second equation:

{y = -2 x - 2

{10 x - 3 (-2 x - 2) = -2

10 x - 3 (-2 x - 2) = (6 x + 6) + 10 x = 16 x + 6:

{y = -2 x - 2

{16 x + 6 = -2

in the second equation, look to solve for x:

{y = -2 x - 2

{16 x + 6 = -2

subtract 6 from both sides:

{y = -2 x - 2

{16 x = -8

divide both sides by 16:

{y = -2 x - 2

{x = -1/2

substitute x = -1/2 into the first equation:

{y = -1

{x = -1/2

collect results in alphabetical order:

answer:   {x = -1/2 ,   y = -1