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}`.