two maths qs

Two terms, a and b, are inserted btw 2 and 4√2 such that the 4 consecutive terms form a GP. If a,c and b form anarithmetic sequence, then c=

2, a, b, 4√2 form a GP, the common ratio :
a/2 = b/a = (4√2)/b

a / 2 = b / a
b = a²/2 ...... [1]

a/2 = (4/√2)/b
ab = 8√2 ...... [2]

Put [1] into [2] :
a(a²/2) = 8√2
a³ = 16√2
a³ = (2√2)³
a = 2√2

Put a = 2√2 into [1] :
b = (2√2)²/2
b = 4

a, c, b form an AP, the common difference :
c - a = b - c
2c = a + b
c = (a + b)/2 ...... [3]

Put a =2√2 and b = 4 into [3] :
c = (4 + 2√2)/2
c = 2 + √2


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The graph of y = x2 + kx + 3 intersects with the line y = x - 1 at P(x1,y1) and Q(x2, y2). y1 + y2 =

y = x² + kx + 3 ...... [1]
y = x - 1 ...... [2]

From [2] :
x = y + 1 ...... [3]

Put [3] into [1] :
y = (y + 1)² + k(y + 1) + 3
y = y² + 2y + 1 + ky + k + 3
y² + (k + 1)y + (k + 4) = 0

y1 and y2 are the roots of the above equation.
Sum of the roots :
y1 + y2 = -(k + 1)