CISE 9 Discuss the nature of the roots of the following quadratic equations, without solving the equations: \( \begin{array}{lll}\text { (1) } x^{2}-3 x-10=0 & \text { (2) } x^{2}-2 x+1=0 \\ \text { (4) } x^{2}-x+3=0 & \text { (5) } 2 x^{2}+6=-7 x \\ \text { (7) }(x+2)(x-1)=4 x & \text { (8) } x^{2}+p^{2}=2 p x & \text { (6) } x^{2}-5 x-7=0 \\ \text { (9) } k x^{2}+(4 k-1) x=2-4 k\end{array} \)

To discuss the nature of the roots of the given quadratic equations, we will use the discriminant \( D \) of a quadratic equation in the standard form \( ax^2 + bx + c = 0 \). The discriminant is calculated as: \[ D = b^2 - 4ac \] The nature of the roots can be determined as follows: - If \( D > 0 \): The equation has two distinct real roots. - If \( D = 0 \): The equation has exactly one real root (a repeated root). - If \( D < 0 \): The equation has two complex (non-real) roots. Now, let's analyze each equation step by step. 1. **Equation (1):** \( x^2 - 3x - 10 = 0 \) - Here, \( a = 1 \), \( b = -3 \), \( c = -10 \). - Calculate \( D = (-3)^2 - 4(1)(-10) = 9 + 40 = 49 \). - Since \( D > 0 \), there are two distinct real roots. 2. **Equation (2):** \( x^2 - 2x + 1 = 0 \) - Here, \( a = 1 \), \( b = -2 \), \( c = 1 \). - Calculate \( D = (-2)^2 - 4(1)(1) = 4 - 4 = 0 \). - Since \( D = 0 \), there is exactly one real root. 3. **Equation (3):** \( x^2 - x + 3 = 0 \) - Here, \( a = 1 \), \( b = -1 \), \( c = 3 \). - Calculate \( D = (-1)^2 - 4(1)(3) = 1 - 12 = -11 \). - Since \( D < 0 \), there are two complex roots. 4. **Equation (4):** \( 2x^2 + 6 = -7x \) (Rearranging gives \( 2x^2 + 7x + 6 = 0 \)) - Here, \( a = 2 \), \( b = 7 \), \( c = 6 \). - Calculate \( D = (7)^2 - 4(2)(6) = 49 - 48 = 1 \). - Since \( D > 0 \), there are two distinct real roots. 5. **Equation (5):** \( (x+2)(x-1) = 4x \) (Rearranging gives \( x^2 - 5x + 2 = 0 \)) - Here, \( a = 1 \), \( b = -5 \), \( c = 2 \). - Calculate \( D = (-5)^2 - 4(1)(2) = 25 - 8 = 17 \). - Since \( D > 0 \), there are two distinct real roots. 6. **Equation (6):** \( x^2 - 5x - 7 = 0 \) - Here, \( a = 1 \), \( b = -5 \), \( c = -7 \). - Calculate \( D = (-5)^2 - 4(1)(-7) = 25 + 28 = 53 \). - Since \( D > 0 \), there are two distinct real roots. 7. **Equation (7):** \( (x+2)(x-1) = 4x \) (Rearranging gives \( x^2 - 5x + 2 = 0 \)) - This is the same as Equation (5), so the nature of the roots is the same: two distinct real roots. 8. **Equation (8):** \( x^2 + p^2 = 2px \) (Rearranging gives \( x^2 - 2px + p^2 = 0 \)) - Here, \( a = 1 \), \( b = -2p \), \( c = p^2 \). - Calculate \( D = (-2p)^2 - 4(1)(p^2) = 4p^2 - 4p^2 = 0 \). - Since \( D = 0 \), there is exactly one real root. 9. **Equation (9):** \( kx^2 + (4k-1)x = 2 - 4k \) (Rearranging gives \( kx^2 + (4k-1)x + (4k-2) = 0 \)) - Here, \( a = k \), \( b = 4k - 1 \), \( c = 4k - 2 \). - Calculate \( D = (4k - 1)^2 - 4k(4k - 2) = (16k^2 - 8k + 1) - (16k^2 - 8k) = 1 \). - Since \( D > 0 \), there are two distinct real roots. ### Summary of the Nature of Roots: 1. Two distinct real roots. 2. One real root (repeated). 3. Two complex roots. 4. Two distinct real roots. 5. Two distinct real roots. 6. Two distinct real roots. 7. Two distinct real roots. 8. One real root (repeated). 9. Two distinct real roots.