Question 3 Solve the equation: \( 2 x^{2}+7 x+3=0 \)

To solve the quadratic equation: \[ 2x^{2} + 7x + 3 = 0 \] we can use the **quadratic formula**, which is a standard method for solving any quadratic equation of the form \( ax^{2} + bx + c = 0 \). The quadratic formula is: \[ x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a} \] **Step 1: Identify the coefficients** For the given equation: \[ a = 2,\quad b = 7,\quad c = 3 \] **Step 2: Calculate the discriminant (\( \Delta \))** The discriminant determines the nature of the roots and is given by: \[ \Delta = b^{2} - 4ac \] Plugging in the values: \[ \Delta = 7^{2} - 4 \times 2 \times 3 = 49 - 24 = 25 \] Since the discriminant is positive (\( \Delta > 0 \)), there are two distinct real roots. **Step 3: Apply the quadratic formula** \[ x = \frac{-b \pm \sqrt{\Delta}}{2a} = \frac{-7 \pm \sqrt{25}}{2 \times 2} = \frac{-7 \pm 5}{4} \] **Step 4: Compute both solutions** 1. **First Root (\( x_{1} \))** \[ x_{1} = \frac{-7 + 5}{4} = \frac{-2}{4} = -\frac{1}{2} \] 2. **Second Root (\( x_{2} \))** \[ x_{2} = \frac{-7 - 5}{4} = \frac{-12}{4} = -3 \] **Step 5: Final Answer** The solutions to the equation \( 2x^{2} + 7x + 3 = 0 \) are: \[ x = -\frac{1}{2} \quad \text{and} \quad x = -3 \] --- **Alternative Method: Factoring** Sometimes, quadratic equations can be factored into the product of two binomials. Let's attempt to factor the given equation. \[ 2x^{2} + 7x + 3 = 0 \] We look for two numbers that multiply to \( a \times c = 2 \times 3 = 6 \) and add up to \( b = 7 \). These numbers are 6 and 1. Rewrite the middle term using these numbers: \[ 2x^{2} + 6x + x + 3 = 0 \] Factor by grouping: \[ 2x(x + 3) + 1(x + 3) = 0 \] \[ (2x + 1)(x + 3) = 0 \] Set each factor equal to zero: 1. \( 2x + 1 = 0 \) → \( x = -\frac{1}{2} \) 2. \( x + 3 = 0 \) → \( x = -3 \) This confirms the solutions obtained using the quadratic formula.