Question

In x^2-bx+c=0,if one root is twice of the other then prove that 9c=2b^2​

Answer 1

Answer :

9c = 2b²

Step-by-step explanation :

Given polynomial,

x² - bx + c = 0

⇒ x² coefficient = 1

⇒ x coefficient = -b

⇒ constant term = c

Given,

one root is twice the other.

Let the one root be "a"

the other root = 2a

we know that,

Sum of the roots = -(x coefficient)/x² coefficient

   a + 2a = -(-b)/1

      3a = b

        a = b/3

Product of the roots = constant term/x² coefficient

    (a)(2a) = c/1

       2a² = c

     2(b/3)² = c    [ a = b/3 ]

     2(b²/9) = c

       2b² = 9c

         

Hence proved!