Q)  Find the zeroes of the quadratic polynomial 2 x² – (1 + 2√2) x + √2 and verify the relationship between zeroes and coefficients of the polynomial.

Q28 – Sample Question Paper – Set 1 – Maths Standard – CBSE 2026

Ans: In the given polynomial equation, to find zeroes, we will start with f(x) = 0

Therefore, 2 x² – (1 + 2√2) x + √2 = 0

Step 1: Let’s start calculating the zeroes of the polynomial:

∵ 2 x² – (1 + 2√2) x + √2 = 0

∴ 2 x² – x – 2√2 x + √2 = 0

∴ x (2 x – 1) – √2 (2 x – 1) = 0

∴ (x – √2) (2 x – 1) = 0

∴  x = √2 and x = 1 / 2

∴ the value of given polynomial will be zero for x = √2 and x = 1 / 2

Therefore, the zeroes of 2 x² – (1 + 2√2) x + √2 are √2 and 1 / 2.

Step 2: Next we have to verify the relationship between the zeroes and the coefficients of the polynomial.

To do this, we need to find the sum of zeroes and the product of zeroes

We know that, if α and β be the zeros of the polynomial, then

Sum of zeroes, α + β = – b / a

and Product of Zeroes, α × β = c / a

We will find the values of both sides and if these are matched, the relationship between the zeroes and the coefficients will get verified.

Since, we have already calculated values of the zeroes of polynomial, let’s calculate values of the coefficients now.

Step 3: When we compare polynomial 2 x² – (1 + 2√2) x + √2 = 0 with standard quadratic equation ax² + b x + c = 0, we get

a = 2, b = – (1 + 2√2) and c = √2

Step 3: Let’s start verifying the relationships one by one:

First, we take relationship 1 for sum of zeroes: α + β = – b / a

In LHS: α + β = √2 + 1 / 2 = (1 + 2√2 ) / 2

and RHS: – b / a = – [- (1 + 2√2)] / (2) = (1 + 2√2 ) / 2

Since LHS = RHS

Hence, the relation of sum of the zeros (α + β = – b / a ) is verified.

Next, we take relationship 2 for Product of zeroes: α × β = c / a

In LHS: α × β = (√2) x (1 / 2) = √2  / 2 = 1 / √2

and RHS: c / a = √2 / 2 = 1/ √2

Since LHS = RHS

Hence, the relationship of product of zeros (α × β = c / a) is also verified.

Thus, √2 and 1 / 2 are the zeroes of the polynomial.

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