Question

please solve 1(ii) and (iii) question step by step

Answer 1

(1) 

Given a + b = 9 and ab = 20.

We know that a^2 + b^2 = (a + b)^2 - 2ab

                                          = (9)^2 - 2(20)

                                          = 81 - 40

                                          = 41.


(2) 

Given p - q = 6   ------ (1)

           p + q = 14  ----- (2)


On solving (1) & (2), we get

p - q = 6

p + q = 14

-----------------

2p = 20

p = 10.


Substitute p = 10 in (2), we get

p + q = 14

10 + q = 14

q = 4.


Now,

p^2 + q^2 = (10)^2 + (4)^2

                 = 116.



(3) 

Given m + n = 8   -------- (1)

           m - n = 2    ------------- (2) 


On solving (1) & (2), we get

m + n = 8

m - n = 2

----------------

2m = 10

m = 5.


Substitute m = 5 in (2), we get

m + n = 8

 5 + n = 8

n = 3.

Now,

mn =  5 * 3

      = 15.


Hope this helps!

Answer 2

a^2+b^2 when a+b=9. ab=20

Use formula

(a+b)^2=a^2+b^2+2ab
9^2=a^2+b^2+2×20
81=a^2+b^2+40
81-40=a^2+b^2
a^2+b^2=41




p^2+q^2 if p-q=6 p+q=14

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

(14)^2=p^2+q^2+2pq
196=p^2+q^2+2pq
-2pq=p^2+q^2-196

(p-q)^2=p^2+q^2-2pq
6^2=p^2+q^2+p^2+q^2-196
36+196=2(p^2+q^2)
232=2(p^2+q^2)
116=(p^2+q^2)



mn if m+n=8 m-n=2

(m+n)^2=m^2+n^2+2mn

8^2=m^2+n^2+2mn
64-2mn=m^2+n^2

(m-n)^2=m^2+n^2-2mn
2^2=m^2+n^2-2mn
4=m^2+n^2-2mn
4=64-2mn-2mn
4-64=-4mn
-60=-4mn
15=mn