Question

find the value of a,if
pqa=(3p+q)^2-(3p-q)^2

Answer 1

pqa=(3p+q)²-(3p-q)²

a=9p²+6pq+q²-(9p²-6pq+q²)/pq

a=9p²+6pq+q²-9p²+6pq-q²/pq

a=12pq/pq

a=pq(12)/pq

a=12

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Answer 2

Answer:

r^2 = s^2

<=> ((3p + q)/2)^2 = (p – q)^2

<=> 9*p^2 + 6*p*q + q^2 = 4*p^2 - 8*p*q + 4*q^2

<=> 5*p^2 + 14*p*q - 3*q^2 = 0

From here, I set p = k*q where k is a real number to obtain an expression such that (something) * q^2 = 0.

5*k^2*q^2 + 14*k*q^2 - 3*q^2 = 0

<=> (5*k^2 + 14*k - 3)*q^2 = 0

=> 5*k^2 + 14*k - 3 = 0

-3 is a root...

So we have,

5*(k+3)*(k-1/5) = 0

Finally,

p could be -3*q or q/5

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