Question

p+q=6,pq=3 and p>q a.p-q=? b.p^3-q^3-5(p^2-q^2) =? c.p^5+q^5=4806​

Answer 1

Answer:

Step-by-step explanation:

a)

$(p-q)^2= p^2+q^2 - 2pq\\=p^2+q^2+2pq-4pq\\= (p+q)^2-4pq\\= 6^2-4\cdot 3\\= 36 - 12 = 24\\or, p-q = \sqrt{24}~~~ (~\because~p>q \implies p-q>0~)$

b)

$p^3 - q^3 - 5(p^2-q^2)\\=(p-q)(p^2+q^2+pq) - 5(p-q)(p+q)\\= (p-q)((p+q)^2 - pq - 5(p+q))\\=\sqrt{24}(36 - 3 - 5\cdot 6)\\=3\sqrt{24}$

c)

$p^5+q^5\\= (p^4+q^4)(p+q) - pq(p^3+q^3)\\= 6((p^2+q^2)^2 -2 (pq)^2) - 3(p+q)(p^2+q^2-pq)\\= 6(((p+q)^2-2pq)^2 -2 (pq)^2) - 3(p+q)((p+q)^2 -3pq)\\= 6((36-6)^2-2\cdot 9) - 3\cdot 6(36 - 9)\\= 6(900 - 18) - 18\cdot 27\\=4806$