We have seen that a quadratic polynomial is of the form:

$$p\left( x \right):a{x^2} + bx + c,\,\,a \ne 0,$$

We assume that a, b and c are real numbers. In general, this polynomial has two zeroes. For example, the polynomial $$p\left( x \right):{x^2} - 3x + 2$$ has the zeroes $$x = 1,\,\,2$$.

For some quadratic polynomials, the two zeroes might be equal. For example, the polynomial$$p\left( x \right):{x^2} - 4x + 4$$ can be rewritten as $$p\left( x \right):{\left( {x - 2} \right)^2}$$. Thus, we can say that this polynomial has the two zeroes: $$x = 2,\,\,2,$$ which happen to be identical.

There might also be quadratic polynomials which have no real zeroes. Consider the polynomial $$p\left( x \right):{x^2} + 1$$. For no real value of x can this polynomial take zero value, which means that this has no real zeroes. Note that this does not mean that the polynomial does not have any zeroes. Of course this polynomial has (two) zeroes – however, those zeroes are just not real numbers.

Therefore, remember that a quadratic polynomial (we assume that the coefficients are real) will always have two zeroes, but the nature of the zeroes depends on the coefficients:

1. The two zeroes might be real and distinct.

2. The two zeroes might be real and identical.

3. The two zeroes might not be real numbers. In this case, they will be complex numbers.

Example 1: Consider the following quadratic polynomial:

$$p\left( x \right):2{x^2} - 6x + 4$$

Which of the following are zeroes of this polynomial?

(A) $$x = 1$$                                       (B) $$x = 2$$

(C) $$x = 3$$                                       (D) $$x = 4$$

Solution: The correct options are (A) and (B). We have:

$$\begin{array}{l}p\left( 1 \right) = 2{\left( 1 \right)^2} - 6\left( 1 \right) + 4 = 0\\p\left( 2 \right) = 2{\left( 2 \right)^2} - 6\left( 2 \right) + 4 = 0\end{array}$$

Since a quadratic polynomial cannot have more than two zeroes, we do not even need to calculate the values of the polynomial for the last two options.

Example 2: Consider the following quadratic polynomial:

$$p\left( x \right):{\left( {x - 6} \right)^2} + 7$$

This polynomial will have two real and distinct zeroes. Is this true or false?

Solution: The given statement is false. Since the polynomial is the sum of a square term (which is always positive) and a positive integer, it will never become zero. In fact, the minimum value of this polynomial will be:

${p_{\min }} = p\left( 6 \right) = {\left( {6 - 6} \right)^2} + 7 = 7$

The minimum value is attained when the square term attains its minimum value, which is 0. This happens when x is equal to 6. Thus, we conclude that the given polynomial does not have real zeroes.

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