Example 1.3.1 Discuss the nature of the roots of the quadratic equation (a^(2)+b^(2))x^(2)-3(a-b)x+9//2=0,quad a+b!=0

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Example 1.3.1 Discuss the nature of the roots of the quadratic equation  (a^(2)+b^(2))x^(2)-3(a-b)x+9//2=0,quad a+b!=0

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Identify coefficients: Identify the coefficients of the quadratic equation. The quadratic equation is given by (a^2 + b^2)x^2 - 3(a - b)x + 9/2 = 0. We can compare this with the standard form ax^2 + bx + c = 0 to find the coefficients. Here, a = a^2 + b^2, b = -3(a - b), and c = 9/2.Calculate discriminant: Calculate the discriminant of the quadratic equation. The discriminant of a quadratic equation ax^2 + bx + c = 0 is given by D = b^2 - 4ac. We will calculate the discriminant for our equation using the identified coefficients. D = (-3(a - b))^2 - 4(a^2 + b^2)(9/2)Simplify discriminant: Simplify the discriminant. D = 9(a - b)^2 - 4(a^2 + b^2)(9/2) D = 9(a^2 - 2ab + b^2) - 2(2a^2 + 2b^2)(9) D = 9a^2 - 18ab + 9b^2 - 18a^2 - 18b^2 D = -9a^2 + 18ab - 9b^2Factor common term: Factor out the common term. D = -9(a^2 - 2ab + b^2) D = -9(a - b)^2Analyze roots: Analyze the discriminant to discuss the nature of the roots. Since D = -9(a - b)^2 and a square is always non-negative, (a - b)^2 ≥ 0. Therefore, D ≤ 0. If D < 0, the quadratic equation has two complex roots. If D = 0, the quadratic equation has one real repeated root. Since we have D = -9(a - b)^2 and it is given that a + b ≠ 0, which does not affect the sign of D, we can conclude that D < 0.Conclude root nature: Conclude the nature of the roots based on the discriminant. Because D < 0, the roots of the quadratic equation are complex and not real. They will be in the form of a + bi and a - bi, where i is the imaginary unit.

$\newline$ Discuss the nature of the roots of the quadratic equation $\newline$ $(a^{2}+b^{2})x^{2}-3(a-b)x+\frac{9}{2}=0,\quad a+b\neq 0$

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$\newline$ Discuss the nature of the roots of the quadratic equation $\newline$ $(a^{2}+b^{2})x^{2}-3(a-b)x+\frac{9}{2}=0,\quad a+b\neq 0$