Assertion (A): The polynomial $x^{2} + 1$ has no real zeros. Reason (R): A quadratic polynomial whose discriminant is negative has no real zeros. Which is correct?

Class 10PolynomialsAlgebra — Polynomials

ABoth A and R are true and R correctly explains A
BBoth A and R are true but R does not explain A
CA is true but R is false
DA is false but R is true

Step-by-Step Solution

For

x^{2} + 1, D = 0 - 4 (1) (1) = - 4 < 0

, so it has no real zeros. R states the general rule and correctly explains A.

Answer: A) Both A and R are true and R correctly explains A

Key idea: Geometric meaning of zeros

Zeros of

p (x)

are exactly the **

x

-coordinates where the graph

y = p (x)

meets the

x

-axis**.

- **Linear**

y

= ax +

b

: graph is a straight line, crosses

x

-axis ONCE → exactly 1 zero.
- **Quadratic**

y = a x^{2} + b x + c

: graph is a parabola. Possible cases:
- Crosses

x

-axis at 2 points → 2 distinct real zeros.
- Touches

x

-axis at 1 point → 1 repeated zero (vertex on

x

-axis).
- Does not meet

x

-axis → no real zeros.
- **Cubic**

y = a x^{3}

+ …: shape is a curve that crosses 1, 2, or 3 times → 1, 2, or 3 real zeros.

So 'find the zeros' is the algebraic version of 'find where the graph crosses the

x

-axis'.

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Assertion (A): The polynomial x2+1 has no real zeros. Reason (R): A quadratic polynomial whose discriminant is negative has no real zeros. Which is correct?

Step-by-Step Solution

Key idea: Geometric meaning of zeros

Master this chapter in 30 minutes a day.

More Class 10 Polynomials practice

Assertion (A): The polynomial $x^{2} + 1$ has no real zeros. Reason (R): A quadratic polynomial whose discriminant is negative has no real zeros. Which is correct?