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kakasveta [241]

3 years ago

9

the equation x^3-3x=1 has a solution between 1.8 and 1.9. use a trial and improvement method to find this solution. give your an

swer correct to 2 decimal places.

1 answer:

Reika [66]3 years ago

4 0

Answer:

The solution to the equation x³ - 3x = 1 is given as :

1.88

Step-by-step explanation:

In the above question, we are given the equation

x³ - 3x = 1 ... Equation 1

and we are told the answer lies between 1.8 - 1.9

Using the trial and improvement method approximating to 2 decimal places

a. First trial we use 1.84 for x in equation 1

1.84³ - 3×1.84 = 0.71

This first trial using 1.84 too small

b. Second trial we use 1.86 for x in equation 1

1.86³ - 3×1.86 = 0.86

This second trial using 1.86 as well is too small

c. For our third trial we use 1.89 for x in equation 1

1.89³ - 3×1.89 = 1.08

This third trial using 1.89 is large. Our target is 1.00

d. Our fourth trial we use 1.88 for x in equation 1

1.88³ - 3×1.88 = 1.004

Approximately to 2 decimal places.

We have = 1.00

Hence the solution to the equation

x³ - 3x = 1 after using the trial and improvement method is 1.88

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Step-by-step explanation:

Given

See attachment for complete question

Required

Match equivalent expressions

Solving (a):

$3 + 12 + 48 + 192 + 768$

The expression can be written as:

$3 \to 3*4^{0$ --- 0

$12 \to 3 * 4^{1$ ---- 1

$48 \to 3 * 4^{2$ --- 2

$192 \to 3 * 4^{3$ ---- 3

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For the nth term, the expression is:

$Term = 3 * 4^{n$ ---- n

So, the summation is:

$3 + 12 + 48 + 192 + 768 = \sum\limits^4_{n=0} 3 * 4^n$

Solving (b):

$4 + 32 + 256 + 2048 + 16384$

The expression can be written as:

$4 \to 4 * 8^0$ --- 0

$32 \to 4 * 8^1$ ---- 1

$256 \to 4 * 8^2$ --- 2

$2048 \to 4 * 8^3$ ---- 3

$16384 \to 4 * 8^4$ ---- 4

For the nth term, the expression is:

$Term \to 4 * 8^n$ ---- n

So, the summation is:

$4 + 32 + 256 + 2048 + 16384 = \sum\limits^4_{n=0} 4 * 8^n$

Solving (c):

$2 + 6 + 18 + 54 + 162$

The expression can be written as:

$2 \to 2 * 3^0$ --- 0

$6 \to 2 * 3^1$ ---- 1

$18 \to 2 * 3^2$ --- 2

$54 \to 2 * 3^3$ ---- 3

$162 \to 2 * 3^4$ ---- 4

For the nth term, the expression is:

$Term \to 2 * 3^n$ ---- n

So, the summation is:

$2 + 6 + 18 + 54 + 162 = \sum\limits^4_{n=0} 2* 3^n$

Solving (d):

$3 + 15 + 75 + 375 + 1875$

The expression can be written as:

$3 \to 3 * 5^0$ --- 0

$15 \to 3 * 5^1$ ---- 1

$75 \to 3 * 5^2$ --- 2

$375 \to 3 * 5^3$ ---- 3

$1875 \to 3 * 5^4$ ---- 4

For the nth term, the expression is:

$Term \to 3 * 5^n$ ---- n

So, the summation is:

$3 + 15 + 75 + 375 + 1875 = \sum\limits^4_{n=0} 3* 5^n$

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