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:

(i) 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 =

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

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

(ii) Now, we need to verify the relationship between the zeroes and the coefficients of the polynomial.

Step 2: Let’s compare polynomial 2 x² – (1 + 2 √2) x + √2 = 0 with standard quadratic equation a x² + b x + c = 0, we get

a = 2, b = – (1 + 2 √2) and c = √2

Also, We calculated zeroes of the polynomial 2 x² – (1 + 2√2) x + √2 in step 1 as: √2 and

let’s consider: α = √2  and β =

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

First, we take relationship 1 for sum of zeroes: α + β =

In LHS: α + β

= √2 +  =

and RHS:

=  =

Since LHS = RHS, the relation of sum of the zeros (α + β = ) is verified.

Step 4: we take relationship 2 for Product of zeroes, i.e. α × β = c / a

In LHS: α × β

=  

=  =

and RHS:

=  =

Since LHS = RHS, the relationship of product of zeros (α . β = ) is also verified.

Therefore, √2 and are the zeroes of the polynomial. The relationship between the zeroes and coefficients is verified as α + β = ) and α . β =.

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