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Q1(ii):
Which of the following expressions are polynomials in one variable and which are not? State reasons for your answer. (ii) $y^2 + \sqrt{2}$

Solution :

Step 1: Theoretical Foundation of Polynomials

To determine whether a given algebraic expression is a polynomial in one variable, we must evaluate it against two strict algebraic criteria:

  • Single Variable Condition: The expression must contain exactly one distinct variable (e.g., only $x$, only $y$, or only $z$).
  • Non-Negative Integer Exponent Condition: [Per the fundamental definition of polynomials], the exponent of the variable in every term must be a whole number (i.e., $0, 1, 2, 3, \dots$). Fractional, negative, or irrational exponents on the variable disqualify the expression from being a polynomial.

Step 2: Structural Analysis of the Given Expression

The given mathematical expression is:

$y^2 + \sqrt{2}$

We can rewrite this expression in its standard canonical form by explicitly showing the variable in the constant term. Any non-zero constant $c$ can be written as $c \cdot y^0$ because $y^0 = 1$ (for $y \neq 0$).

$y^2 + \sqrt{2} \cdot y^0$

y f(y) Vertex: (0, √2) f(y) = y² + √2

Visual representation of the polynomial function showing a continuous, smooth parabolic curve characteristic of quadratic polynomials.

Step 3: Evaluating the Variable Condition

Observing the expression $y^2 + \sqrt{2}$, the only alphabetical symbol representing an unknown quantity is $y$. There are no other variables (such as $x$ or $z$) present in the expression. Therefore, the expression satisfies the condition of being in one variable.

Step 4: Evaluating the Exponent Condition

We must now inspect the exponent of the variable $y$ in each term:

Term Variable Part Exponent Is it a Whole Number?
First Term: $y^2$ $y^2$ $2$ Yes
Second Term: $\sqrt{2}$ $y^0$ $0$ Yes

Note: The presence of $\sqrt{2}$ does not violate the polynomial definition. The restriction of having non-negative integer powers applies only to the variables, not to the numerical coefficients or constants. $\sqrt{2}$ is simply a real number acting as the constant term.

Step 5: Logical Conclusion

Since the expression contains exactly one variable ($y$) and all exponents of this variable ($2$ and $0$) are non-negative integers (whole numbers), it perfectly satisfies all algebraic axioms defining a polynomial in one variable.

Final Solution: Yes, the expression $y^2 + \sqrt{2}$ is a polynomial in one variable. The reason is that it contains only one variable ($y$), and the exponent of the variable in every term is a whole number.


More Questions from Class 9 Mathematics Polynomials EXERCISE 2.1


CBSE Solutions for Class 9 Mathematics Polynomials


Chapters in CBSE - Class 9 Mathematics


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