Examining the roots of a quadratic equation means to see the type of its roots i.e., whether they are real or imaginary, rational or irrational, equal or unequal.
The nature of the roots of a quadratic equation depends entirely on the value of its discriminant b\(^{2}\)  4ac.
In a quadratic equation ax\(^{2}\) + bx + c = 0, a ≠ 0 the coefficients a, b and c are real. We know, the roots (solution) of the equation ax\(^{2}\) + bx + c = 0 are given by x = \(\frac{b \pm \sqrt{b^{2}  4ac}}{2a}\).
1. If b\(^{2}\)  4ac = 0 then the roots will be x = \(\frac{b ± 0}{2a}\) = \(\frac{b  0}{2a}\), \(\frac{b + 0}{2a}\) = \(\frac{b}{2a}\), \(\frac{b}{2a}\).
Clearly, \(\frac{b}{2a}\) is a real number because b and a are real.
Thus, the roots of the equation ax\(^{2}\) + bx + c = 0 are real and equal if b\(^{2}\) – 4ac = 0.
2. If b\(^{2}\)  4ac > 0 then \(\sqrt{b^{2}  4ac}\) will be real and nonzero. As a result, the roots of the equation ax\(^{2}\) + bx + c = 0 will be real and unequal (distinct) if b\(^{2}\)  4ac > 0.
3. If b\(^{2}\)  4ac < 0, then \(\sqrt{b^{2}  4ac}\) will not be real because \((\sqrt{b^{2}  4ac})^{2}\) = b\(^{2}\)  4ac < 0 and square of a real number always positive.
Thus, the roots of the equation ax\(^{2}\) + bx + c = 0 are not real if b\(^{2}\)  4ac < 0.
As the value of b\(^{2}\)  4ac determines the nature of roots (solution), b\(^{2}\)  4ac is called the discriminant of the quadratic equation.
Definition of discriminant: For the quadratic equation ax\(^{2}\) + bx + c =0, a ≠ 0; the expression b\(^{2}\)  4ac is called discriminant and is, in general, denoted by the letter ‘D’.
Thus, discriminant D = b\(^{2}\)  4ac
Note:
Discriminant of ax\(^{2}\) + bx + c = 0 
Nature of roots of ax\(^{2}\) + bx + c = 0 
Value of the roots of ax\(^{2}\) + bx + c = 0 
b\(^{2}\)  4ac = 0 
Real and equal 
 \(\frac{b}{2a}\), \(\frac{b}{2a}\) 
b\(^{2}\)  4ac > 0 
Real and unequal 
\(\frac{b \pm \sqrt{b^{2}  4ac}}{2a}\) 
b\(^{2}\)  4ac < 0 
Not real 
No real value 
When a quadratic equation has two real and equal roots we say that the equation has only one real solution.
Solved examples to examine the nature of roots of a quadratic equation:
1. Prove that the equation 3x\(^{2}\) + 4x + 6 = 0 has no real roots.
Solution:
Here, a = 3, b = 4, c = 6.
So, the discriminant = b\(^{2}\)  4ac
= 4\(^{2}\)  4 ∙ 3 ∙ 6 = 36  72 = 56 < 0.
Therefore, the roots of the given equation are not real.
2. Find the value of ‘p’, if the roots of the following quadratic equation are equal (p  3)x\(^{2}\) + 6x + 9 = 0.
Solution:
For the equation (p  3)x\(^{2}\) + 6x + 9 = 0;
a = p  3, b = 6 and c = 9.
Since, the roots are equal
Therefore, b\(^{2}\)  4ac = 0
⟹ (6)\(^{2}\)  4(p  3) × 9 = 0
⟹ 36  36p + 108 = 0
⟹ 144  36p = 0
⟹ 36p =  144
⟹ p = \(\frac{144}{36}\)
⟹ p = 4
Therefore, the value of p = 4.
3. Without solving the equation 6x\(^{2}\)  7x + 2 = 0, discuss the nature of its roots.
Solution:
Comparing 6x\(^{2}\)  7x + 2 = 0 with ax\(^{2}\) + bx + c = 0 we have a = 6, b = 7, c = 2.
Therefore, discriminant = b\(^{2}\) – 4ac = (7)\(^{2}\)  4 ∙ 6 ∙ 2 = 49  48 = 1 > 0.
Therefore, the roots (solution) are real and unequal.
Note: Let a, b and c be rational numbers in the equation ax\(^{2}\) + bx + c = 0 and its discriminant b\(^{2}\)  4ac > 0.
If b\(^{2}\)  4ac is a perfect square of a rational number then \(\sqrt{b^{2}  4ac}\) will be a rational number. So, the solutions x = \(\frac{b \pm \sqrt{b^{2}  4ac}}{2a}\) will be rational numbers. But if b\(^{2}\) – 4ac is not a perfect square then \(\sqrt{b^{2}  4ac}\) will be an irrational numberand as a result the solutions x = \(\frac{b \pm \sqrt{b^{2}  4ac}}{2a}\) will be irrational numbers. In the above example we found that the discriminant b\(^{2}\) – 4ac = 1 > 0 and 1 is a perfect square (1)\(^{2}\). Also 6, 7 and 2 are rational numbers. So, the roots of 6x\(^{2}\) – 7x + 2 = 0 are rational and unequal numbers.
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