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# Solve the quadraic equation:

## $$2x^2+4x+39=0$$

Since the discriminant $$\Delta <0$$, the quadratic equation has two conjugate complex roots.

$$\Delta=-296$$

$$\begin{cases} x_1=-1+\dfrac{\sqrt{74}}{2}i \\ x_2=-1-\dfrac{\sqrt{74}}{2}i \end{cases}$$

In decimal notation,

$$\begin{cases} x_1=-1+4.3011626335213i \\ x_2=-1-4.3011626335213i \end{cases}$$

Detailed Steps on Solution

## Solve the quadratic equation: $$2x² + 4x + 39 = 0$$

Given $$a =2, b=4, c=39$$,

### 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{-4\pm\sqrt{4^2-4\cdot 2\cdot 39}}{2 \cdot 2}\\ & =\dfrac{-4\pm2\sqrt{-74}}{4}\\ & =-1\pm\dfrac{\sqrt{74}}{2}i\\ \end{aligned}

Since the discriminat is less than zero, we get two complex roots:

That is,

$$\begin{cases} x_1 =-1+\dfrac{\sqrt{74}}{2}i \\ x_2=-1-\dfrac{\sqrt{74}}{2}i \end{cases}$$

### 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,

$$2x² + 4x + 39 = 0$$

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

$$x^2+2x+\dfrac{39}{2}=0$$

Move the constant term $$\dfrac{39}{2}$$ to the right hand side. Then its sign becomes negative.

$$x^2+2x=-\dfrac{39}{2}$$

Add square of the half of $$2$$, the coefficient of the linear term to both sides.

$$x^2+2x+\Big(1\Big)^2=-\dfrac{39}{2}+\Big(1\Big)^2$$

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

$$\Big(x+1\Big)^2=-\dfrac{37}{2}$$

Taking square roots on both sides of above equation gives

$$\sqrt{\Big(x+1\Big)^2}=\pm\sqrt{-\dfrac{37}{2}}$$

Since the left hand side is square root of a perfect square, we can get rid of radical. On the right hand side, the constant $$-\dfrac{37}{2}$$ is less than zero. There's no real roots for square root. However, we can take out the minus by introducing imaginary number. Then,

$$x+1=\pm\sqrt{\dfrac{37}{2}}i$$

Move the constant $$1$$ to the right hand side. Then we get,

$$x_1 = -1+\sqrt{\dfrac{37}{2}}i$$

$$x_2 = -1-\sqrt{\dfrac{37}{2}}i$$

### 3. The vertex of the function $$f(x) = 2x² + 4x + 39$$

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, $$a=2$$, $$b=4$$ and $$c=39$$. 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{-4}{2\cdot 2}\\ & =-1\\ \end{aligned}

\begin{aligned} \\y_{min}&=\dfrac{4ac-b^2}{4a}\\ & =\dfrac{4\cdot 2\cdot39-4^2}{4\cdot2^2}\\ & =37\\ \end{aligned}

So the coordinates for the vertex of the quadrautic function are $$\Big(-1,37\Big)$$

### 4. Reflecting function of $$f(x) = 2x² + 4x + 39$$

$$f(x)= a\Big( x+\dfrac{b}{2a}\Big)^2 +\dfrac{4ac-b^2}{4a}$$

Let $$g(x)$$ be the reflecting function of $$f(x)$$ with respect to the tangent to its vertex.

Then,

$$g(x)=-a\Big(x+\dfrac{b}{2a}\Big)^2+\dfrac{4ac-b^2}{4a}$$

Expanding and simplifing gives

$$g(x)=-ax^2-bx+c-\dfrac{b^2}{2a}$$

Find the Zeros for $$g(x)$$

$$x = \dfrac{b\pm\sqrt{b^2-4(-a)(c-\dfrac{b^2}{2a} ) }}{-2a}$$

$$=\dfrac{b\pm\sqrt{4ac-b^2}}{-2a}$$

It shows the discriminant of $$g(x)$$ is always greater than 0 if the discriminant of $$f(x)$$ is less than $$0$$. Therefore, $$g(x)$$ has two distinct real roots.

Substituting the values of $$a,b,c$$ gives the reflecting fucntion of $$f(x) = 2x² + 4x + 39$$ as below,

$$g(x)=-x^2-4x+35$$

Find the roots by using the root formula.

\begin{aligned} \\x&=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\\ & =\dfrac{-\Big(-4\Big)\pm\sqrt{\Big(-4\Big)^2-4\cdot \Big(-2\Big)\cdot 35}}{2\cdot \Big(-2\Big)}\\ & =\dfrac{4\pm\sqrt{16-(-280)}}{-4}\\ & =\dfrac{4\pm\sqrt{296}}{-4}\\ & =-1\mp\dfrac{\sqrt{74}}{2}\\ \end{aligned}

Therefore, the foots for the reflecting function are

$$x_1=-1-\dfrac{\sqrt{74}}{2}$$

$$x_2=-1+\dfrac{\sqrt{74}}{2}$$

in decimal notation,

$$x_1=-5.3011626335213$$

$$x_2=3.3011626335213$$

### 5. Graph for the function $$f(x) = 2x² + 4x + 39$$

Since the discriminat is less than zero, the curve of the cubic function $$f(x) = 2x² + 4x + 39$$ has no intersection points with horizontal axis. However, the complex roots could be represented by ploting the circle which is centered at the midpoint of two roots of its reflection equation. The two intersection points of the circle with axis x represent the roots of the reflection equation. And the two intersection points with the axis y represent the two complex roots of the equation if we take the coordinate system as a complex plane.