(a) Derive the smallest equation whose coefficients are integers and which has roots of \(\frac{1}{2}\) and -7.
(b) Three years ago, a father was four times as old as his daughter is now. The product of their present ages is 430. Calculate the ages of the father and daughter.
(a) Roots of the equation = \(\frac{1}{2}\) and -7
\(\therefore (x - \frac{1}{2})(x + 7) = 0\)
\(x^{2} - \frac{1}{2} x + 7x - \frac{7}{2} = 0\)
\(x^{2} + \frac{13}{2} x - \frac{7}{2} = 0\)
Multiply through by 2 to get integers as coefficients,
\(2x^{2} + 13x - 7 = 0\)
(b) Let the father and daughter's present ages be r and t respectively.
Three years ago, father = (r - 3) years
\((r - 3) = 4t \implies r = (4t + 3) ... (1)\)
\(r \times t = 430 ... (2)\)
Substitute (1) in (2),
\((4t + 3) \times t = 430 \implies 4t^{2} + 3t = 430\)
\(\implies 4t^{2} + 3t - 430 = 0\)
\(4t^{2} - 40t + 43t - 430 = 0\)
\(4t(t - 10) + 43(t - 10) = 0\)
\(\therefore (4t + 43)(t - 10) = 0\)
\(t = \frac{-43}{4}; t = 10\)
Since t is age, it cannot be negative. Hence, t = 10 years.
Recall : \(r = 4t + 3\)
\(\therefore r = 4(10) + 3 = 40 + 3 = 43\)
Therefore, the father's present age is 43 years while the daughter's age is 10 years.
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