Substitute the values into the formula for compound interest.

Formula for compound interest

A = P {(1 + \frac{r}{100})}^{n}

.
A = final amount = 206.46
P = initial amount = 200
r = interest rate = r
n = number of years = 2.

Substitute the values into the formula to form an equation.

200 {(1 + \frac{r}{100})}^{2} = 206.46

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Divide both sides by 200.

{(1 + \frac{r}{100})}^{2} = \frac{206.46}{200} {(1 + \frac{r}{100})}^{2} = 1.0323

Expand the brackets by multiplying every term in the first bracket by every term in the second bracket.

\begin{array}{rcl} (1 + \frac{r}{100}) (1 + \frac{r}{100}) & = & 1.0323 \\ 1 (1 + \frac{r}{100}) + \frac{r}{100} (1 + \frac{r}{100}) & = & 1.0323 \\ 1 + \frac{r}{100} + \frac{r}{100} + {(\frac{r}{100})}^{2} & = & 1.0323 \\ 1 + \frac{2 r}{100} + {(\frac{r}{100})}^{2} & = & 1.0323 \end{array}

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Set equal to zero and rearrange.

\begin{array}{rcl} 1 + \frac{2 r}{100} + {(\frac{r}{100})}^{2} - 1.0323 & = & 0 \\ {(\frac{r}{100})}^{2} + \frac{2 r}{100} - 0.0323 & = & 0 \end{array}

Remove the fractions by multiplying every term by 100² .

\begin{array}{rcl} {(\frac{r}{100})}^{2} {(100)}^{2} + \frac{2 r}{100} {(100)}^{2} - 0.0323 {(100)}^{2} & = & 0 \\ \frac{10000 r^{2}}{10000} + \frac{200000 r}{100} - 323 & = & 0 \end{array}

r^{2} + 200 r - 323 = 0

[1]

Cyclone Bank	Tornado Bank
Invest 25 000 dollars 4.5% compound interest per year for 3 years	Invest 25 000 dollars Receive 1150 dollars interest each year for 3 years

Compound Interest & Depreciation (Edexcel IGCSE Maths)

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