# Solve the quadraic equation:

## $$2x^2+19x+31=0 $$

**Quick Answer**

Since the discriminant $$\Delta >0$$, the quadratic equation has two distinct real roots.

$$ \Delta=113$$

$$\begin{cases} x_1=-\dfrac{19}{4}+\dfrac{\sqrt{113}}{4} \\ x_2=-\dfrac{19}{4}-\dfrac{\sqrt{113}}{4} \end{cases}$$

In decimal notation,

$$\begin{cases} x_1=-2.0924635468163 \\ x_2=-7.4075364531837 \end{cases}$$

**Detailed Steps on Solution **

## Solve the quadratic equation: $$2x² + 19x + 31 = 0$$

Given $$a =2, b=19, c=31$$,

### 1. Use the quadratic root

Use the root solution formula for a quadratic equation, the roots of the equation are given as

$$\begin{aligned} \\x&=\dfrac{-b\pm\sqrt{b^2-4ac} }{2a}\\ & =\dfrac{-19\pm\sqrt{19^2-4\cdot 2\cdot 31}}{2 \cdot 2}\\ & =\dfrac{-19\pm\sqrt{113}}{4}\\ & =-\dfrac{19}{4}\pm\dfrac{\sqrt{113}}{4}\\ \end{aligned}$$

Since the discriminat is greater than zero, we get two real roots:That is,

$$\begin{cases} x_1 =-\dfrac{19}{4}+\dfrac{\sqrt{113}}{4} \\ x_2=-\dfrac{19}{4}-\dfrac{\sqrt{113}}{4} \end{cases}$$

### 2. Completing the square method

The idea of completing the square is to transform a quadratic equation to the form of a perfect square equals to a constant, which could be positive or minus. And then solve the equation by taking square root on both sides. For the quadratic equation,

$$2x² + 19x + 31 = 0$$

divide each term by $$2$$ to make the coefficient of the leading term $$1$$.

$$x^2+\dfrac{19}{2}x+\dfrac{31}{2}=0$$

Move the constant term $$\dfrac{31}{2}$$ to the right hand side. Then its sign becomes negative.

$$x^2+\dfrac{19}{2}x=-\dfrac{31}{2}$$

Add square of the half of $$\dfrac{19}{2}$$, the coefficient of the linear term to both sides.

$$x^2+\dfrac{19}{2}x+\Big(\dfrac{19}{4}\Big)^2=-\dfrac{31}{2}+\Big(\dfrac{19}{4}\Big)^2$$

Convert the trinomial to the form of perfect square on the left hand side. Sum on the left hand side. Then,

$$\Big(x+\dfrac{19}{4}\Big)^2=\dfrac{113}{16}$$

Taking square roots on both sides of above equation gives

$$\sqrt{\Big(x+\dfrac{19}{4}\Big)^2}=\pm\sqrt{\dfrac{113}{16}}$$

Since the left hand side is square root of a perfect square, we can get rid of radical. Then,

$$x+\dfrac{19}{4}=\pm\dfrac{\sqrt{113}}{4}$$

Move the constant $$\dfrac{19}{4}$$ to the right hand side. Then we get,

$$x_1 = -\dfrac{9}{2}$$

$$x_2 = -5$$

### 3. The vertex of the function $$f(x) = 2x² + 19x + 31$$

The vertex of a quadratic function could be determined by completing the square method to transform the quadratic function from general form to vertex form.

For the general form of a quadratic function, we can do the following transformation.

$$\begin{aligned} \\f(x)&=ax^2+bx+c\\ & =a(x^2+\dfrac{b}{a}x)+c\\ & =a\Big[x^2+\dfrac{b}{a}x+\Big(\dfrac{b}{2a}\Big)^2-\Big(\dfrac{b}{2a}\Big)^2\Big]+c\\ & =a\Big(x+\dfrac{b}{2a}\Big)^2+c-\dfrac{b^2}{4a}\\ & =a\Big(x+\dfrac{b}{2a}\Big)^2+\dfrac{4ac-b^2}{4a}\\ \end{aligned}$$

Therefore, the corrdinates of the vertex is, $$\Big(\dfrac{-b}{2a},\dfrac{4ac-b^2}{4a}\Big)$$

Here we have, $$a=2$$, $$b=19$$ and $$c=31$$. Since $$a >0$$, the curve of the function has the vertex at its lowest point. Substitute them to the vertex formula.

$$\begin{aligned} \\x&=\dfrac{-b}{2a}\\ & =\dfrac{-19}{2\cdot 2}\\ & =-\dfrac{19}{4}\\ \end{aligned}$$

$$\begin{aligned} \\y_{min}&=\dfrac{4ac-b^2}{4a}\\ & =\dfrac{4\cdot 2\cdot31-19^2}{4\cdot2^2}\\ & =-\dfrac{113}{8}\\ \end{aligned}$$

So the coordinates for the vertex of the quadrautic function are $$\Big(-\dfrac{19}{4},-\dfrac{113}{8}\Big)$$

### 4. Graph for the function $$f(x) = 2x² + 19x + 31$$

Since the discriminat is greater than zero, the curve of the cubic function $$f(x) = 2x² + 19x + 31$$ has two intersection point with the x-axis