Which statement about p(x) = x² − 6x + 9 is correct?

It has exactly one distinct zero, x = 3.. x² − 6x + 9 = (x − 3)², so the only value making p(x) = 0 is x = 3 (a repeated zero). A quadratic has AT MOST two distinct real zeros — here there is just one.

Which statement about $p (x) = x^{2} - 6 x + 9$ is correct?

Class 9PolynomialsPolynomials

AIt has exactly one distinct zero, $x = 3$ .
BIt has two distinct zeros, $x = 3$ and $x = - 3$ .
CIt has no real zero.
DIt has two distinct zeros, $x = 1$ and $x = 9$ .

Step-by-Step Solution

x^{2} - 6 x + 9 = (x - 3)^{2}

, so the only value making

p (x) = 0

x = 3

(a repeated zero). A quadratic has AT MOST two distinct real zeros — here there is just one.

Answer: A) It has exactly one distinct zero,

x = 3

Key idea: Finding zeros (roots)

**Linear**: ax +

b = 0

→

x = - b / a

. One zero.

Example:

3 x - 12 = 0

→

x = 4

.

**Quadratic**: solve by factorisation, or quadratic formula (Class 10).

Example:

x^{2} - 5 x + 6 = 0

. Factor:

(x - 2) (x - 3) = 0

→

x = 2

x = 3

. Two zeros.

**General principle**: a polynomial of degree n has AT MOST n real zeros.

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Which statement about p(x)=x2−6x+9 is correct?

Step-by-Step Solution

Key idea: Finding zeros (roots)

Master this chapter in 30 minutes a day.

More Class 9 Polynomials practice

Which statement about $p (x) = x^{2} - 6 x + 9$ is correct?