Question

1. If x + y = 12 and xy = 32, Find the value of x2 + y2.
2. If 3x + 2y = 12 and xy = 6, find the value of 9x2 + 4y2.
3. Write the following cubes in the expanded form:
(i) (3a + 4b)3
(ii) (5p – 3q)3

Answer 1

1)X+Y=12
Y=12-X

XY=32
X(12-X)=32
x^2-12x+32=0
factors of X are 8 and 4
if X=8 then y =4
x^2+y^2 = 8^2+ 4^2. =64+16=80

if X=4 then y =8
x^2+y^2 = 4^2+ 8^2. =16+64=80

2)3x+2y=12
2y=12-3x
y=(12-3x)/2
substitute in XY=6
after factorisation value of x is 2
value of y is 3

value of 9x2 + 4y2 = 9*4+4*9=72

3) (3a+4b)^3
expand using (a+b)^3 formula
(a+b)^3 = a3 + b3 + 3a2b + 3ab2

(a-b)^3 = a3 - b3 - 3a2b + 3ab2

Answer 2

Answer:

1st part

given =.x+y = 12

xy=32

find $x^{2}+y^{2}=?$

x+y = 12 (squaring on both side)

$x^{2}+y^{2}+ 2xy = 144 \\ x^{2}+y^{2}+2×32 = 144 \\ x^{2}+y^{2}=.144-2×32 \\ x^{2}+y^{2}=80$

2nd part see attachment