Problem 8 Proof

A student claims that

Given any value of $p \in Z$ , there is some $x \in R$ such that
$p = \frac{x^{2} - 3 x}{x - 4}, x \neq 4$

Hint

For (a), set $p = 4$ and multiply both sides by $x - 4$ . Use the discriminant.

Hint

For (b), multiply both sides by $x - 4$ to get

x^{2} - 3 x = p (x - 4)

Now collect this into a quadratic in terms of $x$ .

This equation is unsolvable exactly for those values of $p$ such that the discriminant is negative.

Solution

By rearranging, we have

\begin{aligned} p & = \frac{x^{2} - 3 x}{x - 4} \\ p (x - 4) & = x^{2} - 3 x \\ p x - 4 p & = x^{2} - 3 x \\ x^{2} - (p + 3) x + 4 p & = 0 \end{aligned}

When $p = 4$ this becomes
$x^{2} - 7 x + 16 = 0$
The discriminant is $(- 7)^{2} - 4 (16) = - 15$ . Since the discriminant is negative, this equation has no solutions. Hence, the original equation has no solutions in $x$ when $p = 4$ . This disproves the student's claim.
The equation $x^{2} - (p + 3) x + 4 p = 0$ has no solutions when it discriminant is negative
$\begin{aligned} (p + 3)^{2} - 4 (4 p) & < 0 \\ p^{2} + 6 p + 9 - 16 p & < 0 \\ p^{2} - 10 p + 9 & < 0 \\ (p - 9) (p - 1) & < 0 \end{aligned}$
We make a sketch to help
From here, we see that the claim fails for values of $p$ where $1 < p < 9$ so
$p \in {2, 3, 4, 5, 6, 7, 8}$
are the values of $p$ where the claim fails.

Problem 8 Proof ​