Question

solve solve the following equation given that the roots of each are in GP
$54 {x}^{3} - 39 {x}^{2} - 2x + 16 = 0$
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Answer 1

Answer :

8/9 , -2/3 , 1/2

Note :

★ The general form of a cubic equation is ; ax³ + bx² + cx + d = 0 .

★ Let α , ß and γ be the roots of the cubic equation ax³ + bx² + cx + d = 0 , then

• Sum of roots , α + ß + γ = -b/a

• Sum of roots taking two at a time ,

αß + ßγ + αγ = c/a

• Product of roots , αßγ = -d/a

★ GP(Geometric Progression) : A sequence is said to be in GP if the ratio between the consecutive terms are equal .

★ If a1 , a2 , a3 , . . . , an are in GP , then

a2/a1 = a3/a2 = a4/a3 = . . .

★ Three numbers in GP can be given as ;

a/r , a , ar .

★ Four numbers in GP can be given as ;

a/r³ , a/r , ar , ar³ .

Solution :

Here ,

The given cubic equation is ;

54x³ - 39x² - 2x + 16 = 0 .

Now ,

Comparing the given cubic equation with the general cubic equation

ax³ + bx² + cx + d = 0 , we have ;

a = 54

b = -39

c = -2

d = 16

Now ,

=> Product of roots = -d/a

=> a/r × a × ar = -16/54

=> a³ = -8/27

=> a³ = (-2/3)³

=> a = -2/3

Now ,

=> Sum of roots = -b/a

=> a/r + a + ar = -(-39)/54

=> a(1/r + 1 + r) = 39/54

=> a(1 + r + r²)/r = 13/18

=> 18a(r² + r + 1) = 13r

=> 18×(-2/3)×(r² + r + 1) = 13r

=> -12(r² + r + 1) = 13r

=> -12r² - 12r - 12 = 13r

=> 12r² + 12r + 13r + 12 = 0

=> 12r² + 25r + 12 = 0

=> 12r² + 9r + 16r + 12 = 0

=> 3r(4r + 3) + 4(4r + 3) = 0

=> (4r + 3)(3r + 4) = 0

=> r = -3/4 , -4/3

★ If a = -2/3 and r = -3/4 , then the roots of the given cubic equation will be ;

• a/r = (-2/3) / (-3/4) = (-2/3) × (-4/3) = 8/9

• a = -2/3

• ar = (-2/3) × (-3/4) = 1/2

★ If a = -2/3 and r = -4/3 , then the roots of the given cubic equation will be ;

• a/r = (-2/3) / (-4/3) = (-2/3) × (-3/4) = 1/2

• a = -2/3

• ar = (-2/3) × (-4/3) = 8/9

Hence ,

The roots of the given cubic equation are ; 8/9 , -2/3 , 1/2 .