# How do you complete the square to solve 2x² + 3x -5=0?

Jul 2, 2015

$x = 1$ or $x = - \frac{5}{2}$

#### Explanation:

You follow a series of steps to complete the square.

Step 1. Write your equation in standard form.

$2 {x}^{2} + 3 x - 5 = 0$

Step 2. Move the constant to the right hand side of the equation.

Add $5$ to each side of the equation.

$2 {x}^{2} + 3 x - 5 + 5 = 0 + 5$

$2 {x}^{2} + 3 x = 5$

Step 3. Divide both sides of the equation by the coefficient of ${x}^{2}$.

${x}^{2} + \frac{3}{2} x = \frac{5}{2}$

Step 4. Square the coefficient of x and divide by 4.

${\left(\frac{3}{2}\right)}^{2} / 4 = \frac{\frac{9}{4}}{4} = \frac{9}{16}$

Step 5. Add the result to each side.

${x}^{2} + \frac{3}{2} x + \frac{9}{16} = \frac{5}{2} + \frac{9}{16}$

${x}^{2} + \frac{3}{2} x + \frac{9}{16} = \frac{40}{16} + \frac{9}{16}$

${x}^{2} + \frac{3}{2} x + \frac{9}{16} = \frac{49}{16}$

Step 6. Take the square root of each side.

x+3/4 = ±7/4

${x}_{1} + \frac{3}{4} = + \frac{7}{4}$

x_1 = 7/4 – 3/4 = (7-3)/4 = 4/4 = 1

${x}_{1} = 1$

${x}_{2} + \frac{3}{4} = - \frac{7}{4}$

x_2 = -7/4 – 3/4 = (-7-3)/4 = (-10)/4 = -5/2

So $x = 1$ or $x = - \frac{5}{2}$

Check: Substitute the values of $x$ back into the quadratic.

(a) $x = 1$

2x^2 + 3x -5 = 2(1)^2 + 3(1) -5 = 2+ 3 – 5 = 0.

(b) $x = - \frac{5}{2}$

2x^2 + 3x -5 = 2(-5/2)^2 + 3(-5/2) -5 = 2(25/4) – 15/2 -5 = 25/2 -15/2-10/2 = (25-15-10)/2 = 0

Jul 2, 2015

Solve: $y = 2 {x}^{2} + 3 x - 5 = 0$

#### Explanation:

To solve this type of quadratic equations, use the Shortcut. It can save you a lot of work and time.
(a + b + c = 0) -> 2 real roots: 1 and $\left(\frac{c}{a} = - \frac{5}{2}\right)$