Nature of the roots of the Equation ax^{2} + bx + c = 0
The two roots of the Quadratic equation ax^{2} + bx + c = 0 are:
The expression b^{2} – 4ac which appear under radical sign is called the Discriminant (Disc.) of the quadratic equation. i.e., Disc = b^{2} – 4ac
The expression b^{2} – 4ac discriminates the nature of the roots, whether they are real, rational, irrational or imaginary. There are three possibilities.
Conditionb^{2} – 4ac > 0 |
Natureis a perfect square, the roots are real, rational and unequal. is not a perfect square, then roots are real, irrational and unequal. |
b^{2} – 4ac = 0 |
then roots will be real, equal and rational. |
b^{2} – 4ac < 0 |
then roots will be imaginary and unequal. |
Example 1 :
Find the nature of the roots of the following equations
2x^{2} + 3x + 1 = 0
Solution :
Comparing the given quadratic equation with ax^{2} + bx + c = 0, we get
a = 2, b = 3 and c = 1
b^{2} - 4ac = 3^{2} - 4(2) (1)
= 9-8
= 1 > 0
Since it is perfect square, roots are real, rational and unequal.
Example 2 :
Find the nature of the roots of the following equations
6x^{2} = 7x + 5
Solution :
6x^{2} = 7x + 5
Converting into standard form, we get
6x^{2} - 7x - 5 = 0
Comparing the given quadratic equation with ax^{2} + bx + c = 0, we get
a = 6, b = -7 and c = -5
b^{2} - 4ac = (-7)^{2} - 4(6) (-5)
= (49 + 120)
= 169 > 0
Since it is perfect square, roots are real, rational and unequal.
Example 3 :
Find the nature of the roots of the following equations
3x^{2} + 7x - 2 = 0
Solution :
3x^{2} + 7x - 2 = 0
Comparing the given quadratic equation with ax^{2} + bx + c = 0, we get
a = 3, b = 7 and c = -2
b^{2} - 4ac = 7^{2} - 4(3) (-2)
= (49 + 24)
= 73 > 0
Since it is not perfect square, roots are real, irrational and unequal.
Example 4 :
Find the nature of the roots of the following equations
√2x^{2} + 3x - √8 = 0
Solution :
√2x^{2} + 3x - √8 = 0
Comparing the given quadratic equation with ax^{2} + bx + c = 0, we get
a = √2, b = 3 and c = -√8
b^{2} - 4ac = 3^{2} - 4(√2) (-√8)
= 9 + 16
= 25 > 0
Since it is perfect square, roots are real, rational and unequal.
Example 5 :
Show that the roots of the following equations are rational
a(b – c)x^{2} + b(c – a)x + c(a – b) = 0
Solution :
a = a(b - c), b = b(c - a) and c = c(a - b)
b^{2} - 4ac = [b(c - a)]^{2} - 4a(b - c) c(a - b)
= b^{2}(c - a)^{2} - 4ac(b - c)(a - b)
= b^{2}(c^{2} - 2ac + a^{2}) - 4ac(ab - b^{2} - ac + bc)
= b^{2} c^{2} - 2acb^{2} + a^{2}b^{2} - 4a^{2}bc + 4acb^{2} + 4a^{2}c^{2} - 4abc^{2}
= a^{2}b^{2 }+ 4a^{2}c^{2 }+ b^{2} c^{2} - 4a^{2}bc - 4abc^{2 }+2 acb^{2}
= (ab)^{2} + (-2ac)^{2} + (bc)^{2} - 4a^{2}bc - 4abc^{2 }+2 acb^{2}
= (ab - 2ac + bc)^{2}
Since it is perfect square, the given quadratic equation will have rational roots.
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