How do you solve using the completing the square method # 2x^2+3x-5=0#?

Question

Genesis Allen · Accepted Answer

#color(brown)(x = +-7/4 - 3/4 = 1, -5/2# #2x^2 + 3x - 5 = 0# #cancel2(x^2 + (3/2)x - 5/2 )= 0# #(x^2 + (2 * (3/4) * x ) = 5/2# Add #(3/4)^2# to both sides. #x^2 + (2 * (3/4) * x) + (3/4)^2 = 5/2 + (3/4)^2 = 49/16# #(x + 3/4)^2 = (sqrt(49/16))^2 = (7/4)^2# #x + 3/4 = +- 7/4# #color(brown)(x = +-7/4 - 3/4 = 1, -5/2#

Abigail Allen · Answer

undefined To solve the equation $2x^2 + 3x - 5 = 0$ using the completing the square method:

1. Move the constant term to the other side of the equation: $2x^2 + 3x = 5$.
2. Divide all terms by the coefficient of $x^2$ to make the coefficient 1: $x^2 + \frac{3}{2}x = \frac{5}{2}$.
3. Add $(\frac{3}{4})^2$ to both sides to complete the square: $x^2 + \frac{3}{2}x + (\frac{3}{4})^2 = \frac{5}{2} + (\frac{3}{4})^2$.
4. Factor the perfect square trinomial on the left side and simplify the right side: $(x + \frac{3}{4})^2 = \frac{25}{4}$.
5. Take the square root of both sides: $x + \frac{3}{4} = \pm \sqrt{\frac{25}{4}}$.
6. Solve for $x$: $x = -\frac{3}{4} \pm \frac{5}{2}$.
7. Simplify: $x = -\frac{3}{4} \pm \frac{5}{2}$.

So, the solutions are $x = -\frac{3}{4} + \frac{5}{2}$ and $x = -\frac{3}{4} - \frac{5}{2}$.