Solve the quadraic equation:

$$2x^2+33x+49=0 $$

Quick Answer

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

$$ \Delta=697$$

$$\begin{cases} x_1=-\dfrac{33}{4}+\dfrac{\sqrt{697}}{4} \\ x_2=-\dfrac{33}{4}-\dfrac{\sqrt{697}}{4} \end{cases}$$

In decimal notation,

$$\begin{cases} x_1=-1.649810608778 \\ x_2=-14.850189391222 \end{cases}$$

Detailed Steps on Solution

Solve the quadratic equation: $$2x² + 33x + 49 = 0$$

Given $$a =2, b=33, c=49$$,

1. Use the quadratic root

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

$$\begin{aligned} \\x&=\dfrac{-b\pm\sqrt{b^2-4ac} }{2a}\\ & =\dfrac{-33\pm\sqrt{33^2-4\cdot 2\cdot 49}}{2 \cdot 2}\\ & =\dfrac{-33\pm\sqrt{697}}{4}\\ & =-\dfrac{33}{4}\pm\dfrac{\sqrt{697}}{4}\\ \end{aligned}$$

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

That is,

$$\begin{cases} x_1 =-\dfrac{33}{4}+\dfrac{\sqrt{697}}{4} \\ x_2=-\dfrac{33}{4}-\dfrac{\sqrt{697}}{4} \end{cases}$$

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,

$$2x² + 33x + 49 = 0$$

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

$$x^2+\dfrac{33}{2}x+\dfrac{49}{2}=0$$

Move the constant term $$\dfrac{49}{2}$$ to the right hand side. Then its sign becomes negative.

$$x^2+\dfrac{33}{2}x=-\dfrac{49}{2}$$

Add square of the half of $$\dfrac{33}{2}$$, the coefficient of the linear term to both sides.

$$x^2+\dfrac{33}{2}x+\Big(\dfrac{33}{4}\Big)^2=-\dfrac{49}{2}+\Big(\dfrac{33}{4}\Big)^2$$

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

$$\Big(x+\dfrac{33}{4}\Big)^2=\dfrac{697}{16}$$

Taking square roots on both sides of above equation gives

$$\sqrt{\Big(x+\dfrac{33}{4}\Big)^2}=\pm\sqrt{\dfrac{697}{16}}$$

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

$$x+\dfrac{33}{4}=\pm\dfrac{\sqrt{697}}{4}$$

Move the constant $$\dfrac{33}{4}$$ to the right hand side. Then we get,

$$x_1 = -8$$

$$x_2 = -\dfrac{17}{2}$$

3. The vertex of the function $$f(x) = 2x² + 33x + 49$$

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, $$a=2$$, $$b=33$$ and $$c=49$$. Since $$a >0$$, the curve of the function has the vertex at its lowest point. Substitute them to the vertex formula.

$$\begin{aligned} \\x&=\dfrac{-b}{2a}\\ & =\dfrac{-33}{2\cdot 2}\\ & =-\dfrac{33}{4}\\ \end{aligned}$$

$$\begin{aligned} \\y_{min}&=\dfrac{4ac-b^2}{4a}\\ & =\dfrac{4\cdot 2\cdot49-33^2}{4\cdot2^2}\\ & =-\dfrac{697}{8}\\ \end{aligned}$$

So the coordinates for the vertex of the quadrautic function are $$\Big(-\dfrac{33}{4},-\dfrac{697}{8}\Big)$$

4. Graph for the function $$f(x) = 2x² + 33x + 49$$

Since the discriminat is greater than zero, the curve of the cubic function $$f(x) = 2x² + 33x + 49$$ has two intersection point with the x-axis