Question

(2) Determine k such that kx^2 - 4x + 1/2=0 has real and different solutions. (3) (i) What rate of interest compounded annually is needed to make an investment of $6,500 accumulate to an amount of $7166.25 at the end of two years? (ii) How long will it take for the initial amount to double, compounded annually at the rate in (i) above?

Answer 1

2)

The polynomial equation is, $kx\\^2-4x+\frac{1}{2}=0$

Where $a = k, b= -4, c=\frac{1}{2}$

Since the equation has distinct root;

therefore $b\\^2-4ac > 0$

$(-4)\\^2-4\times k \times\frac{1}{2} > 0\\ 16 - 2k>0\\ 16>2k\\ 8>k$ , having two real roots,.

3)

i)

Compounded annually formula is $A = P(1 + r)\\^t$

A = $7166.25

t = 2

P $6500

Therefore,

$7166.25 = 6500(1 + r)\\^2\\ \frac{7166.25}{6500} = (1 + r)\\^2\\ 1.1025 = (1 + r)\\^2\\ \\\sqrt{1.2025} - 1 = r\\ r = 0.050013$

Convert rate of interest as percentage: $R = r \times 100\\$

Rates (R) = 5.002 per year.

ii) The amount double = $2 \times 6500 = \$13000\\$

A = $13000

Principal = $6,500

Rate of interest = 5.002%

t= t

$13000 = 6500(1 + 0.05002)\\ ^ t\\ 2 = (1.05002)\\ ^ t$

Taking log on both sides:

$\log2 = t\log(1.05002)\\ t = \frac{\log 2}{\log 1.05002}\\ 14.201 = t$

Time for amount to be doubled is 14 years and 2 months