Review question

# Two values of $x$ that differ by $5$ satisfy $x^2 - 12x + k = 0$, what is $k$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R5949

## Solution

1. 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 $x = -3$.

1. 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 $k = \dfrac{119}{4}.$