Question

Number 14 part 1 pls​

Answer 1

Given

$7 - px - x^2 = 16 - (q + x)^2$

expanding the right side gives

$7 - px - x^2 = 16 - (q^2 + 2qx + x^2)$

$7 - px - x^2 = 16 - q^2 - 2qx - x^2$

Two polynomials of equal degree are the same if their coefficients are identical. This means

$\begin{cases}16 - q^2 = 7 \\ -2q = -p\end{cases}$

$16 - q^2 = 7 \implies q^2 = 9 \implies q=\pm3$

$-2q = -p \implies p = 2q = \pm6$

Both $p$ and $q$ are positive, so $\boxed{p=6}$ and $\boxed{q=3}$.