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5.6 Chapter 5 Case 1: When the sum and product of the roots of a quadratic equation are both positive, each root is positive. For example, if the sum of the roots is 5 and the product of the roots is 6, the roots are 2 and 3. Case 2: If the sum of the roots is positive and the product of the roots is negative, the root with the greatest degree is is positive and the other roots are negative. For example, if the sum of the roots is 6 and the product of the roots is −16, the roots are +8 and -2. Because 8 × (−2) = −16 and +8 − 2 = 6. The larger root, which is 8, is positive and the smaller root, which is 2, is negative. Case 3: If the sum of the roots is negative and the product of the roots is positive, then both roots are negative. For example, if the sum of the roots is −10 and the product of the roots is 24, the roots are −6 and −4. −6 −4 = −10 and (−6)(−4) = 24. In this case, both roots, −6 and −4, are negative in sign. Case 4: If the sum of the roots is negative and the product of the roots is negative, the larger root is negative and the other roots are positive. For example, if the sum of the roots is −4 and the product of the roots is −21, the roots are −7 and +3. (−7) × 3 = −21 and −7 + 3 = −4. The larger root, which is 7, is negative and the smaller root, which is 3, is positive. The above information can be tabulated as follows: Sign Addition Sign Root of Roots Product + ve Both roots are positive + ve Root + ve One root is positive, the other is negative. −ve −ve The largest root is positive in terms of a number. −ve −ve One root is positive and the other is negative. Larger numerical roots are negative. + ve Both negative roots.
What Is The Discriminant Of 9×2 2 10x
Equations and Quadratic Equations 5.7 Constructing quadratic equations with known roots Let a and b be the roots of a quadratic equation. The quadratic equation can be written as (x − a)(x − b) = 0. ⇒ x2 − (a + b) x + ab = 0 That is, x2 − (sum of roots) x + (product of the roots) = 0 ⇒ x2 − −b x + c = 0. a a Example 5.3 Solve the equation x2 − 11x + 30 = 0 using the formula. Solution The given equation is x2 − 11x + 30 = 0. The root of ax2 + bx + c = 0 is −b ± b2 − 4ac . 2a Here, a = 1, b = −11 and c = 30. That is, x = −(−11) ± (−11)2 − 4(1)(30) = 11 ± 121 − 120 2 × 1 2 x = 11 ± 1 2 x = 11 + 1 and 11 − 1 ⇒ 6 and 5. 2 2 ∴ The roots are 5 and 6. Example 5.4 Find the roots of the equation below: (a) x2 − 13x + 11 = 0 (b) 18×2 − 14x + 17 = 0 (c) 9×2 − 36x + 36 = 0 (d) 3×2 − 5x − 8 = 0 Solution (a) Given x2 − 13x + 11 = 0 .Compare with ax2 + b = 0, we have a = 1, b = −13 and c = 11. Now, b2 − 4ac = (−13)2 − 4(1)(11) = 169 − 44 = 125 > 0 .
Discriminant Of Quadratic Equation 3√(3)x^2 + 10x + √(3) = 0
5.8 Chapter 5 ⇒ b2 − 4ac > 0 and is not a perfect square. ∴ The roots are different and irrational. (b) Since 18×2 − 14x + 17 = 0. Compare this with ax2 + bx + c = 0, a = 18, b = −14 and c = 17. Now, b2 − 4ac = ( − 14)2 − 4 ( 18 )(17) = 196 − 1224 = −1028 < 0 ⇒ b2 − 4ac 0 ⇒ b2 − 4ac > 0 and it is a perfect square. ∴ The roots are rational and specific. Example 5.5 1 1 α2 β2 If a, b are the roots of the equation x2 − lx + m = 0, find the value of + in terms of l and m. Solution With roots a, b of x2 − lx + m = 0. ⇒ Sum of roots = α + β = −( −l ) = l (1) 1 ⇒ Product of roots = ab = m = m (2) 1
Quadratic equations and equations 5.9 Now, 1 + 1 = α2 + β2 α2 β2 α 2β 2 = (α + β )2 − 2(αβ ) (αβ )2 Substituting the values a + b and ab instead, in the equation above we get, 1 + 1 = l2 − 2m α2 β2 m2 Value 1 1 l 2 − 2m . α2 β2 m2 ∴ + = Example 5.6 Write a quadratic equation whose roots are 5 and 8 . 2 3 Solution Number of roots = 5 + 8 = 31 . 2 3 6 Product of roots = 5 8 = 20 . 2 3 3 The required quadratic equation is, x2 − (sum of roots)x + (product of roots) = 0. ⇒ x2 − 31 x + 20 = 0 6 ∣ 6 31x + 40 = 0 ∴ A quadratic equation with roots 5 and 8 is 6×2 − 31x + 40 = 0. 2 3 Example 5.7 If one of the roots of the equation, x2 − 11x + (p − 3) = 0 is 3, then find the value of p and also other roots. Solution One of the roots of the equation x2 − 11x + p − 3 = 0 is known to be 3. ⇒ x = 3 satisfies the given equation. ⇒ (3)2 − 11(3) + p − 3 = 0 ⇒ p = 33 + 3 − 9 ⇒ p = 27 ∴ The value of p is 27. Since the sum of the roots of the equation is 11 and one of the roots 3, the other root of the equation is 8.
5.10 Chapter 5 Equations Reducible to Quadratic Form Example 5.8 Solve (x2 − 2x)2 − 23(x2 − 2x) + 120 = 0. Solution Say x2 − 2x = y ⇒ Reduce the given equation to the quadratic equation in y That is, y2 − 23y + 120 = 0 ⇒ y2 − 15y − 8y + 120 = 0 ⇒ y(y − 15) − 8(y − 15) = 0 − (y − 15) = 0 − (y − − − ) 0 ⇒ y − 8 = 0 (or) y − 15 = 0 ⇒ y = 8 (or) y = 15 But x2 − 2x = y When y = 8, x2 − 2x = 8 ⇒ x2 − 2x − − 8 = . x2 − 4x + 2x − 8 = 0 ⇒ x(x − 4) + 2(x − 4) = 0 ⇒ (x + 2)(x − 4) = 0 ⇒ x + 2 = 0 (or) x − 4 = 0 ⇒ x = −2 (or) x = 4 When y = 15, x2 − 2x = 15 ⇒ x2 − 2x − 15 = 0 ⇒ x2 − 5x + 3x − 15 = 0 − − x( ) (x − 5 ) = 0 ⇒ (x − 5)(x + 3) = 0 ⇒ x − 5 = 0 (or) x + 3 = 0 ⇒ x = 5 (or) x = −3 ∴ x
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