The Total solution for NCERT class 6-12
Given that the zeroesof the cubic polynomial x3 – 6x2 + 3x + 10 are of the form a, a + b, a +2b for some real numbers a and b, findthe values of a and b as well as the zeroesof the given polynomial.
Solution:
Given that a, a+b,a+2b are roots of given polynomial x³-6x²+3x+10
Sum of the roots ⇒ a+2b+a+a+b = -coefficient of x²/ coefficientof 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
Substituting the valueof 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