Quadratic Function and Parabola Discriminant: ∆ ax2 + bx + c =0 ax2 + bx + c =0 By Mr Porter

The Discriminant: ∆ From the solution of the general quadratic ax2 + bx + c = 0 using the quadratic formula: The DISCRIMINANT is defined as: ∆ = b2 – 4ac. The roots of the quadratic equation ax2 + bx + c = 0 can be written as: OR

The Discriminant, ∆, and the Roots of a Quadratic. ∆ = b2 – 4ac for ax2 + bx + c = 0, a ≠ 0. There are 4 cases to consider with the discriminant, as shown below. ∆ = b2 – 4ac Case 1: ∆ < 0 Case 2: ∆ = 0 Case 3 and 4: ∆ > 0 TWO real unequal roots b2 – 4ac is a negative number. There are no real solutions to the quadratic equation. The quadratic is positive definite (a > 0) or negative definite (a < 0) b2 – 4ac is ZERO. There is ONE real, rational solution to the quadratic equation. The two solutions are equal. Case 3: ∆ > 0 ∆ = n2 Case 4: ∆ > 0 ∆ ≠ n2 There are TWO real, unequal, rational roots to the quadratic equation. There are TWO real, unequal, irrational roots to the quadratic equation.

GRAPHICAL representation of the Discriminant and the Roots of a Quadratic. ∆ = b2 – 4ac Case 1: ∆ < 0 Case 2: ∆ = 0 Case 3-4: ∆ > 0 TWO real unequal roots Case 3: ∆ > 0 ∆ = n2 Case 4: ∆ > 0 ∆ ≠ n2 Note the number of times the curve cuts the x-axis.

Example 1: Calculate the discriminant for each of the following equations and describe the nature of the roots. a) x2 + 6x + 2 = 0 b) 2x2 + 3x + 4 = 0 a = +1, b = +6 c = +2 a = +2, b = +3 c = +4 The roots of 2x2 + 3x + 4 = 0 are Imaginary (complex numbers) roots. There are no real roots. Since a > 0, the quadratic is positive definite. The roots of x2 + 6x + 2 = 0 are Two real distinct irrational roots.

Example 2: For what values of k will the equation 3x2 – 4x – (1 + k) = 0 have (i) Real and different roots. (ii) Real and equal roots. Solution: For 2 real distinct roots ∆ > 0. Solution: For real and equal roots ∆ = 0. b2 – 4ac > 0, a = 3, b = -4 and c = -(1+k) b2 – 4ac = 0, a = 3, b = -4 and c = -(1+k) (-4)2 – 4(3)(-(1 + k)) > 0 (-4)2 – 4(3)(-(1 + k)) = 0 16 + 12(1 + k) > 0 16 + 12(1 + k) = 0 16 + 12 + 12k > 0 16 + 12 + 12k = 0 28 + 12k > 0 28 + 12k = 0 12k > -28 12k = -28 The quadratic to have real and different roots. The quadratic to have 2 real and equal roots. This could have been stated from part (i).

Example3: For what values of ‘m’ does the equation x2 – 2(m +2)x + 9m = 0 have equal roots? Find these roots for each value of ‘m’. Solution: For real and equal roots ∆ = 0. To find the roots, substitute the 2 values of ‘m’ and solve the resulting quadratics. a = 1, b = -2(m +2), c = 9m x2 – 2(m +2)x + 9m = 0 , m = 1 x2 – 2(1 +2)x + 9(1) = 0 Expand and simplify. x2 – 6x + 9 = 0 Factorise. Expand and simplify. (x – 3)( x – 3) = 0 a double root at x = 3. Factorise. x2 – 2(m +2)x + 9m = 0 , m = 4 x2 – 2(4 +2)x + 9(4) = 0 Expand and simplify. x2 – 12x + 36 = 0 Factorise. (x – 6)( x – 6) = 0 a double root at x = 6.

Example 4: For what value(s) of k will the quadratic expression kx2 – 6x + (6k +3) be a perfect square. A PERFECT SQAURE quadratic has a single (double) real root and ‘a’ and ‘c’ must be square numbers. For real and equal roots ∆ = 0. a = k, b = -6, c = (6k + 3) Now, check Rearrange Factorise. Neither are perfect square numbers. Now, check k = 1 1x2 – 6x + (6 x 1 + 3) = 1x2 – 6x + 9 So, a = 1 and c = 9, Now, we check each value of k! Both are perfect square numbers. Hence, when k = 1, kx2 – 6x + (6k +3) forms a perfect square quadratic.

Example 5: For what values k is the expression x2 – (5k +1)x + 9k2 positive for all values of x. For the expression x2 – (5k +1)x + 9k2 positive for all values of x (Positive Definite) we need to show a > 0 and b2 – 4ac < 0 for the quadratic expression ax2 + bx + c. x2 – (5k +1)x + 9k2 = ax2 + bx + c x2 – (5k +1)x + 9k2 is positive for all values of x, when Step 1: a = 1, that is a > 0 Step 2: b2 – 4ac = [-(5k + 1)]2 – 4(1)(9k2) = [25k2 + 10k + 1)]2 – 36k2 = -11k2 + 10k + 1 Convave down, Factorise. = (-11k + 1)(k + 1) Now, (-11k + 1)(k + 1) < 0 for :

Example 6: (Difficult) Show that the roots of the equation px2 + (p + q)x – (p – q) = 0 are always real, where p and q are real. We need to show that: b2 – 4ac ≥ 0. ∆ = b2 – 4ac a = p, b = (p + q), c = -(p – q) ∆ = (p + q)2 – 4(p)[-(p – q)] ∆ = p2 + 2pq + q2 + 4p2 – 4pq ∆ = [p2 – 2pq + q2] + 4p2 ∆ = (p – q)2 + 4p2 Now, n2 ≥ 0 for all real values of n. Therefore, (p – q)2 ≥ 0 and 4p2 ≥ 0 for all real p and q. So, ∆ ≥ 0 for all real p and q. Therefore , px2 + (p + q)x – (p – q) = 0 must always have real roots for all real p and q.

Example 7: (Difficult) Prove that the roots of the equation 3px2 – 2px – 3qx + 2q = 0 has rational roots if p, q are rational. We need to show that: b2 – 4ac ≥ 0 and ∆ = n2, a squared number. Rearranging the quadratic: 3px2 – 2px – 3qx + 2q = 3px2 – (2p + 3q)x + 2q ∆ = b2 – 4ac a = 3p, b = -(2p + 3q), c = 2q ∆ = [-1 (2p + 3q)]2 – 4(3p)(2q) ∆ = 4p2 + 2(3p)(2q) + 9q2 – 24pq ∆ = 4p2 – 12pq + 9q2 ∆ = (2p – 3q)2 ∆ ≥ 0 and ∆ = n2 for all real p and q. Therefore the quadratic 3px2 – 2px – 3qx + 2q = 0 has real roots.