Find the solution of x²+3x-18=0 by completing square​

To solve the quadratic equation \( x^2 + 3x - 18 = 0 \) by completing the square, follow these steps:1. **Move the constant term to the right side:** \[ x^2 + 3x = 18 \]2. **Complete the square:** To complete the square, take half of the coefficient of \( x \) (which is 3), square it, and add it to both sides of the equation. \[ \text{Half of 3 is } \frac{3}{2} \] \[ \left(\frac{3}{2}\right)^2 = \frac{9}{4} \] Add \(\frac{9}{4}\) to both sides: \[ x^2 + 3x + \frac{9}{4} = 18 + \frac{9}{4} \]3. **Rewrite the left side as a perfect square:** \[ \left(x + \frac{3}{2}\right)^2 = \frac{72}{4} + \frac{9}{4} = \frac{81}{4} \] So, the equation now is: \[ \left(x + \frac{3}{2}\right)^2 = \frac{81}{4} \]4. **Take the square root of both sides:** \[ x + \frac{3}{2} = \pm \frac{\sqrt{81}}{2} = \pm \frac{9}{2} \]5. **Solve for \( x \):** \[ x = -\frac{3}{2} \pm \frac{9}{2} \] This gives two solutions: \[ x_1 = -\frac{3}{2} + \frac{9}{2} = \frac{6}{2} = 3 \] \[ x_2 = -\frac{3}{2} - \frac{9}{2} = -\frac{12}{2} = -6 \]So, the solutions are \( x = 3 \) and \( x = -6 \).

Answered by rhenanuengo | 2024-09-03