If the roots of x^ 3  42x^ 2 + 336x  512 = 0 are in increasing geometric progression, its common ratio is: a) 2 b) 3 c) 4 d) 6
Question
If the roots of $x^3  42x^2 + 336x  512 = 0$ are in increasing geometric progression, its common ratio is:
a) 2 b) 3 c) 4 d) 6
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Answer
To solve for the common ratio ( r ) of the increasing geometric progression for the roots of the given polynomial ( x^3  42x^2 + 336x  512 = 0 ):

Identify the roots in GP: Since the roots are in increasing geometric progression, let's denote them as ( a ), ( ar ), and ( ar^2 ) where ( a ) is the first term and ( r ) is the common ratio.

Sum of roots: According to Vieta's formulas, the sum of the roots of the cubic equation ( x^3 + bx^2 + cx + d = 0 ) is given by ( b ). For our equation: [ a + ar + ar^2 = 42 ] Simplifying, this becomes: [ a(1 + r + r^2) = 42 ]

Product of roots: Similarly, the product of the roots of a cubic equation ( x^3 + bx^2 + cx + d = 0 ) is ( d ). For our equation: [ a \cdot ar \cdot ar^2 = 512 ] Simplifying, this gives: [ a^3 r^3 = 512 ] Taking the cube root of both sides: [ ar = 8 ]

Substitute ( a = \frac{8}{r} ) into the sum equation: [ \frac{8}{r} \left(1 + r + r^2\right) = 42 ] Simplifying further: [ 8 (1 + r + r^2) = 42r ] [ 8 + 8r + 8r^2 = 42r ] Rearranging: [ 8r^2  34r + 8 = 0 ]

Solving the quadratic equation: [ 4r^2  17r + 4 = 0 ] Using the quadratic formula ( r = \frac{b \pm \sqrt{b^2  4ac}}{2a} ): [ r = \frac{17 \pm \sqrt{17^2  4 \cdot 4 \cdot 4}}{2 \cdot 4} ] [ r = \frac{17 \pm \sqrt{225}}{8} ] [ r = \frac{17 \pm 15}{8} ] So, [ r = \frac{32}{8} = 4 \quad \text{or} \quad r = \frac{2}{8} = \frac{1}{4} ]

Choosing the increasing ratio: Since the problem states the roots are in increasing geometric progression, we select ( r = 4 ) (and not ( \frac{1}{4} ) which would indicate a decreasing progression).
Therefore, the common ratio ( r ) is ( \boxed{4} ).
Final answer: c) 4
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