- The equation \(x^3 + ax + b = 0\) is satisfied by the values \(x = 1\) and \(x = 2\). Find the values of \(a\) and \(b\)…
Since \(x = 1\) satisfies the equation \(x^3 + ax + b = 0\), by substituting in this value we find that \[ 1^3 + a \times 1 + b = 0 \implies a + b = - 1. \]
Likewise, since \(x = 2\) satisfies the equation, we also have that \[ 2^3 + a \times 2 + b = 0 \implies 2a + b = -8. \]
We therefore have a pair of simultaneous equations: \[\begin{align*} a + b &= -1, \\ 2a + b &= -8. \end{align*}\] By writing the first equation as \(a = -1 - b\) and substituting this value into the second equation, we are led to a single equation in \(b\), namely \[\begin{align*} 2(-1-b) + b = -8 &\implies -2 - 2b + b = - 8 \\ &\implies b = 6. \end{align*}\]Consequently, \(a = - 1 - b = -7\), and so the original polynomial is \(x^3 - 7x + 6\).
Check: \(2(-7) + 6 = -8\).
…and find also the third value of \(x\).
Since \(x = 1\) and \(x = 2\) are roots of the polynomial \(x^3 - 7x + 6\), we can write \[ x^3 - 7x + 6 = (x-1)(x-2)(x + \beta) \] for some real constant \(\beta\). Since the constant terms have to match, this implies that \[ 6 = (-1) \times (-2) \times \beta \implies \beta = 3. \] That is, \(x^3 - 7x + 6 = (x-1)(x-2)(x+3)\), and the third (and final) root of the polynomial is \(x = -3\).
- The equation \(x^2 - 12x + k = 0\) is satisfied by two values of \(x\) that differ by \(5\). Find these two values of \(x\), and the value of \(k\).
Without any loss of generality, we can call the roots \(\gamma\) and \(\gamma + 5\). So
\[\gamma^2 - 12\gamma + k = 0, (\gamma+5)^2 - 12(\gamma+5)+k=0.\]
Subtracting the first equation from the second, we have \(10\gamma -35 = 0\), and so \(\gamma = \dfrac{7}{2}\).
We have \(k = 12\gamma - \gamma^2\), and so \(k = \dfrac{119}{4}.\)
