Q)  Obtain the zeroes of the polynomial 7 x² + 18 x – 9 Hence, write a polynomial each of whose zeroes is twice the zeroes of given polynomial.

PYQ: Q 27 – CBSE 2025 – Code 30 – Series 5 – Set 1

Ans: 

Step 1: Given polynomial equation 7 x² + 18 x – 9 = 0

Comparing with standard polynomial, a x² + b x + c = 0, we get,

a =  7, b = 18, c = – 9

Since, its given that the roots of the polynomial be α and β.

and we know that sum of roots (α + β) =  – b / a

∴   α + β =  – (18) / (7) = – 18/ 7  …………… (i)

Also, we know that the product of the roots (α x β) = c/ a

∴ α . β = (- 9) / (7) = – 9 / 7 …………. (ii)

Step 2: The zeroes for new polynomial  given as (2 𝛼 , 2 𝛽 ):

∵ Sum of the zeroes of new polynomial = (2 α + 2 β ) = 2 (α + β) 

By transferring values from equations (i), we get:

∴ Sum of the zeroes of new polynomial = 2 (- 18/ 7) = – 36 / 7

Next, Product of the zeroes of new polynomial = (2 α).(2 β) = 4 (α . β)

By transferring values from equations (ii), we get:

∴ Product of the zeroes of new polynomial = 4 (- 9 / 7) = – 36 / 7

Step 3: New quadratic polynomial f(x) = k [x² – (sum of the zeroes) x + (product of the zeroes)]

∴ f(x) = k [x² – (- 36 /7) x + (- 36 / 7)]

For f(x) = 0, since k ≠ 0;  ∴  polynomial x² + (36/ 7) x – 36 /7 = 0

∴ 7 x² + 36 x –  36 = 0

Hence, the required quadratic polynomial is 7 x² + 36 x – 36 = 0.

[Note: For more practice, take another similar question – the link is given below:

Click here for another similar question.]

Please do press “Heart” button if you liked the solution.