Math, asked by Bipinchandra99, 7 months ago

solve the following equation given that each root is in AP
 {x}^{3}  - 3 {x}^{2} - 6x + 8 = 0

Answers

Answered by AlluringNightingale
3

Answer :

-2 , 1 , 4

Solution :

★ 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

★ A.P. (Arithmetic Progression) : A sequence in which the difference between the consecutive terms are equal is said to be in A.P.

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

a2 - a1 = a3 - a2 = a4 - a3 = . . .

★ Three numbers in AP can be given as ;

a - d , a , a + d .

★ Four numbers in AP can be given as ;

a - 3d , a - d , a + d , a + 3d .

Solution :

Here ,

The given cubic equation is ;

x³ - 3x² - 6x + 8 = 0 .

Comparing the given cubic equation with the general cubic equation

ax³ + bx² + cx + d = 0 , we have ;

a = 1

b = -3

c = -6

d = 8

Also ,

It is given that , the roots of the given cubic equation are in AP .

Thus ,

Let the roots of the given cubic equation be ; a - d , a , a + d .

Now ,

=> Sum of roots = -b/a

=> (a - d) + a + (a + d) = -(-3)/1

=> 3a = 3

=> a = 3/3

=> a = 1

Now ,

=> Product of roots = -d/a

=> (a - d)×a×(a + d) = -8/1

=> a(a - d)(a + d) = -8

=> a(a² - d²) = -8

=> 1(1² - d²) = -8

=> 1 - d² = -8

=> 1 + 8 = d²

=> 9 = d²

=> d² = 9

=> d = √9

=> d = ±3

★ If a=1 and d=3 , then the roots of the given cubic equation will be ;

• a - d = 1 - 3 = -2

• a = 1

• a + d = 1 + 3 = 4

★ If a=1 and d=-3 , then the roots of the given cubic equation will be ;

• a - d = 1 - (-3) = 1 + 3 = 4

• a = 1

• a + d = 1 + (-3) = 1 - 3 = -2

Hence ,

The roots of the given cubic equation are ; -2 , 1 , 4 .

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