Question

if p-q=9 and pq=36 evaluate 1)p+q 2) p^2-q^2

Answer 1

Step-by-step explanation:

n:

p-q=9                  

pq=36

i)) given p-q=9

     (p-q)²=9²               (squaring both sides LHS and RHS)

     p²+q²-2pq = 81     

     p²+q²-2×36=81     (substituting value of pq)

     p²+q²-72=81         (shifting -72 from LHS to RHS) 

     p²+q²=153..................................(1)

Now consider the equation

(p+q)²=p²+q²+2pq

(p+q)²=153+(2×36)    (substituting p²+q² from (1)  and pq(given))

(p+q)²=153+72=225

(p+q)²=225

Taking square root on both sides:

√(p+q)²=√225                                ((root of a square no),is the no itself)

   p+q=15 

ii)) now considering p²-q²

p²-q²=(p-q)×(p+q)         (algebraic formula)

p²-q²=(9)×(15)      (p-q=given     :   p+q=found from 1st answer)

p²-q²=135

Answer 2

Answer:

p+q = 15 ; p^2-q^2 = 24

Step-by-step explanation:

p-q = 9      and pq = 36

p-q = 9

squaring both side

(p-q)^2 = 9^2

p^2 + q^2 -2pq = 81

p^2 + q^2 = 81 + 2 * 36                                [pq = 36]

p^2 + q^2 = 153

therefore, value of (p+q)^2

(p+q)^2 = p^2 + q^2 +2pq

(p+q)^2 = 153 + 72                                [p^2 + q^2 = 153]

(p+q) = √225 =15

value of p^2-q^2= (p-q)(p+q)

= 9+15 = 24