(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?
2)
The polynomial equation is,ย "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\\\\\n16 - 2k>0\\\\\n16>2k\\\\\n8>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\\\\\n\\frac{7166.25}{6500} = (1 + r)\\\\^2\\\\\n1.1025 = (1 + r)\\\\^2\\\\\n\n\\\\\\sqrt{1.2025} - 1 = r\\\\\nr = 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\\\\\n2 = (1.05002)\\\\ ^ t"
Taking log on both sides:
"\\log2 = t\\log(1.05002)\\\\\nt = \\frac{\\log 2}{\\log 1.05002}\\\\\n14.201 = t"
Time for amount to be doubled is 14 years and 2 months
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