Academic Inquiry Assistant

Sure! Let's go through the step-by-step solution for the given quadratic equation \( 3x^2 - 5x + 2 = 0 \). ### Step 1: Identify the coefficients For the quadratic equation \( ax^2 + bx + c = 0 \), identify the coefficients \( a \), \( b \), and \( c \): - \( a = 3 \) - \( b = -5 \) - \( c = 2 \) ### Step 2: Calculate the discriminant The discriminant (\( \Delta \)) of a quadratic equation is given by: \[ \Delta = b^2 - 4ac \] Substitute the coefficients: \[ \Delta = (-5)^2 - 4(3)(2) \] \[ \Delta = 25 - 24 \] \[ \Delta = 1 \] ### Step 3: Determine the nature of the roots The discriminant helps determine the nature of the roots of the quadratic equation: - If \( \Delta > 0 \): The equation has two distinct real roots. - If \( \Delta = 0 \): The equation has exactly one real root (a repeated root). - If \( \Delta < 0 \): The equation has two complex conjugate roots. Since \( \Delta = 1 \) is positive, the equation has two distinct real roots. ### Step 4: Calculate the roots The roots of the quadratic equation can be found using the quadratic formula: \[ x = \frac{-b \pm \sqrt{\Delta}}{2a} \] Substitute the coefficients and the value of the discriminant: \[ x = \frac{-(-5) \pm \sqrt{1}}{2(3)} \] \[ x = \frac{5 \pm 1}{6} \] This gives us two roots: \[ x_1 = \frac{5 + 1}{6} = \frac{6}{6} = 1 \] \[ x_2 = \frac{5 - 1}{6} = \frac{4}{6} = \frac{2}{3} \] ### Summary The roots of the quadratic equation \( 3x^2 - 5x + 2 = 0 \) are \( x = 1 \) and \( x = \frac{2}{3} \). ### Significance of the Discriminant - **Positive Discriminant (\( \Delta > 0 \))**: Indicates two distinct real roots. - **Zero Discriminant (\( \Delta = 0 \))**: Indicates exactly one real root (a repeated root). - **Negative Discriminant (\( \Delta < 0 \))**: Indicates two complex conjugate roots. Let's briefly illustrate the other cases for completeness: #### Zero Discriminant Example Consider the equation \( x^2 - 2x + 1 = 0 \): - \( a = 1 \), \( b = -2 \), \( c = 1 \) - \( \Delta = (-2)^2 - 4(1)(1) = 4 - 4 = 0 \) With \( \Delta = 0 \): \[ x = \frac{-(-2) \pm \sqrt{0}}{2(1)} = \frac{2 \pm 0}{2} = 1 \] This equation has one real root: \( x = 1 \). #### Negative Discriminant Example Consider the equation \( x^2 + x + 1 = 0 \): - \( a = 1 \), \( b = 1 \), \( c = 1 \) - \( \Delta = 1^2 - 4(1)(1) = 1 - 4 = -3 \) With \( \Delta = -3 \): \[ x = \frac{-1 \pm \sqrt{-3}}{2(1)} = \frac{-1 \pm \sqrt{3}i}{2} \] This equation has two complex roots: \[ x_1 = \frac{-1 + \sqrt{3}i}{2} \] \[ x_2 = \frac{-1 - \sqrt{3}i}{2} \] These steps illustrate how to determine and find the roots of a quadratic equation based on the discriminant.