We will discuss here about the different cases of discriminant to understand the nature of the roots of a quadratic equation. Show We know that α and β are the roots of the general form of the quadratic equation ax\(^{2}\) + bx + c = 0 (a ≠ 0) .................... (i) then we get α = \(\frac{- b - \sqrt{b^{2} - 4ac}}{2a}\) and β = \(\frac{- b + \sqrt{b^{2} - 4ac}}{2a}\) Here a, b and c are real and rational. Then, the nature of the roots α and β of equation ax\(^{2}\) + bx + c = 0 depends on the quantity or expression i.e., (b\(^{2}\) - 4ac) under the square root sign. Thus the expression (b\(^{2}\) - 4ac) is called the discriminant of the quadratic equation ax\(^{2}\) + bx + c = 0. Generally we denote discriminant of
the quadratic equation by ‘∆ ‘ or ‘D’. Therefore, Discriminant ∆ = b\(^{2}\) - 4ac Depending on the discriminant we shall discuss the following cases about the nature of roots α and β of the quadratic equation ax\(^{2}\) + bx + c = 0. When a, b and c are real numbers, a ≠ 0 Case I: b\(^{2}\) - 4ac > 0 When a, b and c are real numbers, a ≠ 0 and discriminant is positive (i.e., b\(^{2}\) - 4ac > 0), then the roots α and β of the quadratic equation ax\(^{2}\) + bx + c = 0 are real and unequal. Case II: b\(^{2}\) - 4ac = 0 When a, b and c are real numbers, a ≠ 0 and discriminant is zero (i.e., b\(^{2}\) - 4ac = 0), then the roots α and β of the quadratic equation ax\(^{2}\) + bx + c = 0 are real and equal. Case III: b\(^{2}\) - 4ac < 0 When a, b and c are real numbers, a ≠ 0 and discriminant is negative (i.e., b\(^{2}\) - 4ac < 0), then the roots α and β of the quadratic equation ax\(^{2}\) + bx + c = 0 are unequal and imaginary. Here the roots α and β are a pair of the complex conjugates. Case IV: b\(^{2}\) - 4ac > 0 and perfect square When a, b and c are real numbers, a ≠ 0 and discriminant is positive and perfect square, then the roots α and β of the quadratic equation ax\(^{2}\) + bx + c = 0 are real, rational unequal. Case V: b\(^{2}\) - 4ac > 0 and not perfect square When a, b and c are real numbers, a ≠ 0 and discriminant is positive but not a perfect square then the roots of the quadratic equation ax\(^{2}\) + bx + c = 0 are real, irrational and unequal. Here the roots α and β form a pair of irrational conjugates. Case VI: b\(^{2}\) - 4ac is perfect square and a or b is irrational When a, b and c are real numbers, a ≠ 0 and the discriminant is a perfect square but any one of a or b is irrational then the roots of the quadratic equation ax\(^{2}\) + bx + c = 0 are irrational. Notes: (i) From Case I and Case II we conclude that the roots of the quadratic equation ax\(^{2}\) + bx + c = 0 are real when b\(^{2}\) - 4ac ≥ 0 or b\(^{2}\) - 4ac ≮ 0. (ii) From Case I, Case IV and Case V we conclude that the quadratic equation with real coefficient cannot have one real and one imaginary roots; either both the roots are real when b\(^{2}\) - 4ac > 0 or both the roots are imaginary when b\(^{2}\) - 4ac < 0. (iii) From Case IV and Case V we conclude that the quadratic equation with rational coefficient cannot have only one rational and only one irrational roots; either both the roots are rational when b\(^{2}\) - 4ac is a perfect square or both the roots are irrational b\(^{2}\) - 4ac is not a perfect square. Various types of Solved examples on nature of the roots of a quadratic equation: 1. Find the nature of the roots of the equation 3x\(^{2}\) - 10x + 3 = 0 without actually solving them. Solution: Here the coefficients are rational. The discriminant D of the given equation is D = b\(^{2}\) - 4ac = (-10)\(^{2}\) - 4 ∙ 3 ∙ 3 = 100 - 36 = 64 > 0. Clearly, the discriminant of the given quadratic equation is positive and a perfect square. Therefore, the roots of the given quadratic equation are real, rational and unequal. 2. Discuss the nature of the roots of the quadratic equation 2x\(^{2}\) - 8x + 3 = 0. Solution: Here the coefficients are rational. The discriminant D of the given equation is D = b\(^{2}\) - 4ac = (-8)\(^{2}\) - 4 ∙ 2 ∙ 3 = 64 - 24 = 40 > 0. Clearly, the discriminant of the given quadratic equation is positive but not a perfect square. Therefore, the roots of the given quadratic equation are real, irrational and unequal. 3. Find the nature of the roots of the equation x\(^{2}\) - 18x + 81 = 0 without actually solving them. Solution: Here the coefficients are rational. The discriminant D of the given equation is D = b\(^{2}\) - 4ac = (-18)\(^{2}\) - 4 ∙ 1 ∙ 81 = 324 - 324 = 0. Clearly, the discriminant of the given quadratic equation is zero and coefficient of x\(^{2}\) and x are rational. Therefore, the roots of the given quadratic equation are real, rational and equal. 4. Discuss the nature of the roots of the quadratic equation x\(^{2}\) + x + 1 = 0. Solution: Here the coefficients are rational. The discriminant D of the given equation is D = b\(^{2}\) - 4ac = 1\(^{2}\) - 4 ∙ 1 ∙ 1 = 1 - 4 = -3 > 0. Clearly, the discriminant of the given quadratic equation is negative. Therefore, the roots of the given quadratic equation are imaginary and unequal. Or, The roots of the given equation are a pair of complex conjugates. 11 and 12 Grade Math From Nature of the Roots of a Quadratic Equation to HOME PAGE
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We know that a quadratic equation is a second degree polynomial equation of the form ax2 + bx + c = 0, where a, b, c are real numbers, x is the unknown variable and a ≠ 0. For the equation ax2 + bx + c = 0, the discriminant is given by D = b2 – 4ac. It is also denoted by ∆. A quadratic equation has 2 roots. It will be real or imaginary. In this article we discuss the nature of roots depending upon coefficients and discriminant. If α and β are the values of x which satisfy the quadratic equation, α and β are called the roots of the quadratic equation. Roots are given by the equation (-b±√(b2-4ac))/2a. The nature of the roots depends on the discriminant. Nature of Roots depending upon DiscriminantAccording to the value of discriminant, we shall discuss the following cases about the nature of roots. Case 1: D = 0 If the discriminant is equal to zero (b2 – 4ac = 0), a, b, c are real numbers, a≠0, then the roots of the quadratic equation ax2 + bx + c = 0, are real and equal. In this case, the roots are x = -b/2a. The graph of the equation touches the X axis at a single point. Case 2: D > 0 If the discriminant is greater than zero (b2 – 4ac > 0), a, b, c are real numbers, a≠0, then the roots of the quadratic equation ax2 + bx + c = 0, are real and unequal. The graph of the equation touches the X-axis at two different points. Case 3: D < 0 If the discriminant is less than zero (b2 – 4ac < 0), a, b, c are real numbers, a≠0, then the roots of the quadratic equation ax2 + bx + c = 0, are imaginary and unequal. The roots exist in conjugate pairs. The graph of the equation does not touch the X-axis. Case 4: D > 0 and perfect square If D > 0 and a perfect square, then the roots of the quadratic equation are real, unequal and rational. Case 5: D > 0 and not a perfect square If D > 0 and not a perfect square, then the roots of the quadratic equation are real, unequal and irrational. We can summarize all the above cases in the table below.
Nature of Roots depending upon coefficientsDepending upon the nature of the coefficients of the quadratic equation, we can summarize the following.
Bridge Course – Nature of Roots of Quadratic Equations Also Read Quadratic inequalities Solved ExamplesExample 1: The roots of the quadratic equation 3x2-10x+3 = 0 are a) real and equal b) imaginary c) real, unequal and rational d) none of these Solution: Given equation 3x2-10x+3 = 0 Here discriminant, D = b2-4ac => (-10)2 – 4×3×3 = 100 – 36 = 64 D is positive and a perfect square. So the roots of the quadratic equation are real, unequal and rational. Hence option c is the answer. Example 2: Find the value of p if the equation 3x2-18x+p = 0 has real and equal roots. a) 27 b) 18 c) 9 d) none of these Solution: Given 3x2-18x+p = 0 has real and equal roots. => b2-4ac = 0 =>(-18)2-4×3×p = 0 => 324 – 12p = 0 => p = 324/12 = 27 Hence option a is the answer. Example 3: The quadratic equation with real coefficients when one of its root is (3+2i) is Solution: Given one root is 3+2i. Complex roots occur in conjugate pairs. So other root = 3-2i Sum of roots = 6 Product of roots = (3+2i)(3-2i) = 13 Required equation is x2-(Sum)x+Product = 0 => x2-6x+13 = 0 Example 4: Show that the equation 3x2+4x+6 = 0 has no real roots. Solution: Given equation 3x2+4x+6 = 0 Here a = 3, b = 4, c = 6 Discriminant D = b2-4ac => 42-4×3×6 = 16-72 = -56 Since D<0, the roots are imaginary. Hence the equation has no real roots. Video Lesson – Nature of Roots
The discriminant of a quadratic equation is given by D = b2 – 4ac.
If discriminant, D = 0, then the roots are real and equal.
If discriminant, D>0, then the roots are real and unequal.
If discriminant, D<0, then the roots are imaginary and unequal. |
What is the nature of roots of the quadratic equation if the value of its discriminant is negative or less than zero?
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