Let the roots be {a, ar, ar^2}. Then the equation becomes

.. (x -a)*(x -ar)*(x -ar^2) = 0

.. x^3 -a(1 +r +r^2)x^2 +a^2*r*(1 +r +r^2)x -a^3*r^3 = 0

This gives rise to three equations in three unknowns:

.. -14 = -a(1 +r +r^2)

.. k = a^2*r(1 +r +r^2) = (ar)*(14)

.. -64 = -(ar)^3

The last equation tells us

.. ar = 4

.. k = 4*14 = 56

Substituting a = 4/r into the first equation gives

.. 4r^2 -10r +4 = 0

.. 2(r -2)(2r -1) = 0

so (a, r) = (8, 1/2) or (2, 2)

a) roots are 2, 4, 8

b) k = 56

Question:

The roots of the equation x³ − 14x² + kx − 64 = 0 are all real and form a geometric progression. Determine:

a) The roots of the equation;

b) The value of k

Answer:

a)

(x − a) (x − b) (x − c) = x³ − 14x² + kx − 64

x³ − (a+b+c)x² + (ab+ac+bc)x − abc = x³ − 14x² + kx − 64

a + b + c = 14

ab + ac + bc = k

abc = 64

a, b, c in geometric progression:

b = ra

c = r²a

abc = 64

a(ra)(r²a) = 64

(ra)³ = 64

ra = 4

r = 4/a

a + b + c = 14

a + ra + r²a = 14

a + a(4/a) + a(16/a²) = 14

a + 4 + 16/a = 14

a² + 4a + 16 = 14a

a² - 10a + 16 = 0

(a − 2)(a − 8) = 0

a = 2 or 8

r = 4/a = 2 or 1/2

a = 2, b = 4, c = 8

or

a = 8, b = 4, c = 2

Roots: 2, 4, 8

b)

k = ab + ac + bc = 2*4 + 2*8 + 4*8 = 56

Check:

http://www.wolframalpha.com/input/?i=solve+x%C2%B3...