Question

Consider the equation p^2-5p-6-x(p-6)^2=0, where p is a real constant. If p is not equal to 6, then x=

A. (p-2)/(p-6)

B. (p-1)/(p-6)

C. (p+1)/(p-6)

D. (p+2)/(p-6)

Answer 1

Provided that p is a real value not equal to 6, the value of x is (c) $\mathbf{x = \frac{p+1}{p - 6}}$

The equation is given as:

$\mathbf{p^2-5p-6-x(p-6)^2=0}$

Expand

$\mathbf{p^2-6p + p-6-x(p-6)^2=0}$

Factorize

$\mathbf{p(p-6) + 1(p-6)-x(p-6)^2=0}$

Factor out p -6

$\mathbf{(p+1)(p-6)-x(p-6)^2=0}$

Divide through by p - 6

$\mathbf{(p+1)-x(p-6)=0}$

Rewrite as:

$\mathbf{x(p-6) = (p+1)}$

Make x the subject

$\mathbf{x = \frac{p+1}{p - 6}}$

Hence, the value of x is (c) $\mathbf{x = \frac{p+1}{p - 6}}$

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