Calculating Zeroes of a Quadratic Polynomial
Consider the following quadratic polynomial:
\(p\left( x \right):{x^2}  3x + 2\)
To evaluate the zeroes of this polynomial, we can either factorize it, or we can apply the quadratic formula, which will directly give the two roots. We will learn factorization of quadratic polynomials in a subsequent chapter. For now, note that this polynomial can be factorized as
\[p\left( x \right):\left( {x  1} \right)\left( {x  2} \right)\]
This polynomial will therefore be 0 when either of the two factors is 0, or when \(x = 1,\,\,2\). Thus, these are the two zeroes of this quadratic polynomial.
Now, let us talk about the quadratic formula for the zeroes. You will study the justification of this formula later. For now, you should simply remember this formula and learn how it is applied. Consider a general quadratic polynomial:
\[p\left( x \right):a{x^2} + bx + c,\,\,a \ne {\rm{0}}\]
The quadratic formula tells us that the two zeroes of this polynomial are given by
\[x = \frac{{  b \pm \sqrt {{b^2}  4ac} }}{{2a}}\]
One zero corresponds to the plus sign in the formula, and the other corresponds to the minus sign. Let us apply this formula to the polynomial above,
\[p\left( x \right):{x^2}  3x + 2\]
Comparing this with the general form of a quadratic polynomial, we note that
\[a = 1,\,\,\,b =  3,\,\,\,c = 2\]
Thus, applying the quadratic formula, we have
\[\begin{align}&x = \frac{{  \left( {  3} \right) \pm \sqrt {{{\left( {  3} \right)}^2}  4\left( 1 \right)\left( 2 \right)} }}{{2\left( 1 \right)}}\\\,\,\,\,& = \frac{{3 \pm \sqrt {9  8} }}{2} = \frac{{3 \pm 1}}{2} = \frac{4}{2},\frac{2}{2} = 2,1\end{align}\]
These two zeroes are what we obtained by factorization as well.
Example 1: Find the zeroes of
\(p\left( x \right):2{x^2} + 5x + 2\)
Solution: We note that for this case,
\(a = 2,\,\,b = 5,\,\,c = 2\)
Thus, applying the quadratic formula, we have
\[\begin{align}&x = \frac{{  5 \pm \sqrt {{5^2}  4 \times 2 \times 2} }}{{2 \times 2}}\\\,\,\,\, &= \frac{{  5 \pm 3}}{4}\, =  \frac{2}{4},  \frac{8}{4}\, =  \frac{1}{2},  2\end{align}\]
Example 2: Find the zeroes of
\(r\left( x \right):{x^2} + 2x + 1\)
Solution: In this case, \(a = 1,\,\,b = 2,\,\,c = 1\), and so the formula gives
\[\begin{align}&x = \frac{{  2 \pm \sqrt {{2^2}  4 \times 1 \times 1} }}{{2 \times 1}}\\\,\,\,\, &= \frac{{  2 \pm 0}}{2} =  \frac{2}{2},  \frac{2}{2} =  1,  1\end{align}\]
Thus, we have obtained two identical (real roots).
Example 3: Find the zeroes of
\(r\left( x \right):{x^2} + x + 1\)
Solution: We see that for this case, \(a = 1,\,\,b = 1,\,\,c = 1\). Applying the quadratic formula, we have
\[\begin{align}&x = \frac{{  1 \pm \sqrt {{1^2}  4 \times 1 \times 1} }}{{2 \times 1}}\\\,\,\,\,& = \frac{{  1 \pm \sqrt {  3} }}{2}\end{align}\]
We see that by applying our formula, we obtain a negative number under the squareroot sign. Does that mean that the formula has given us a wrong result? No. It means that the zeroes in this case are nonreal (complex).
The nature of the roots is therefore decided by the term under the squareroot sign in the quadratic formula, which is \({b^2}  4ac\). This is called the discriminant of the quadratic polynomial, and will be denoted by D. Three cases are possible:

If D is positive, then the zeroes of the polynomial will be real and distinct.

If D is equal to 0, then the zeroes of the polynomial will be real and identical.

If D is negative, then the zeroes of the polynomial will be nonreal (complex).