NileshKushal
Answered

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number 14 part 1 pls​

Number 14 Part 1 Pls class=

Sagot :

LammettHash

Given

[tex]7 - px - x^2 = 16 - (q + x)^2[/tex]

expanding the right side gives

[tex]7 - px - x^2 = 16 - (q^2 + 2qx + x^2)[/tex]

[tex]7 - px - x^2 = 16 - q^2 - 2qx - x^2[/tex]

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

[tex]\begin{cases}16 - q^2 = 7 \\ -2q = -p\end{cases}[/tex]

[tex]16 - q^2 = 7 \implies q^2 = 9 \implies q=\pm3[/tex]

[tex]-2q = -p \implies p = 2q = \pm6[/tex]

Both [tex]p[/tex] and [tex]q[/tex] are positive, so [tex]\boxed{p=6}[/tex] and [tex]\boxed{q=3}[/tex].

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