Question

14.If the zeroes of the cubic polynomial
x3 - 6x2 + 3x + 10 are of the form
a, a + b and a + 2b for some real numbers a
and b, then find the values of a and b.​

Answer 1

Answer:

Step-by-step explanation:

Given that a, a+b, a+2b are roots of given polynomial x³-6x²+3x+10

From this polynomial,

Sum of the roots ⇒ a+2b+a+a+b = -coefficient of x²/ coefficient of x³

                           ⇒ 3a+3b = -(-6)/1 = 6

                           ⇒ 3(a+b) = 6

                           ⇒ a+b = 2  --------- (1)  b = 2-a

Product of roots ⇒ (a+2b)(a+b)a = -constant/coefficient of x³

                          ⇒ (a+b+b)(a+b)a = -10/1

Placing the value of a+b=2 in it

                         ⇒ (2+b)(2)a = -10

                         ⇒ (2+b)2a = -10

                         ⇒ (2+2-a)2a = -10

                         ⇒ (4-a)2a = -10

                         ⇒ 4a-a² = -5

                         ⇒ a²-4a-5 = 0

                         ⇒ a²-5a+a-5 = 0

                         ⇒ (a-5)(a+1) = 0

                      a-5 = 0      or        a+1 = 0

                      a = 5                     a = -1                

  a = 5, -1 in (1) a+b = 2

When a = 5, 5+b=2 ⇒ b=-3

           a = -1, -1+b=2 ⇒ b= 3

∴ If a=5 then b= -3

            or

   If a= -1 then b=3

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