# Solve the quadraic equation:

## $$x^2-33x-58=0 $$

**Quick Answer**

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

$$ \Delta=1321$$

$$\begin{cases} x_1=\dfrac{33}{2}+\dfrac{\sqrt{1321}}{2} \\ x_2=\dfrac{33}{2}-\dfrac{\sqrt{1321}}{2} \end{cases}$$

In decimal notation,

$$\begin{cases} x_1=34.672781845386 \\ x_2=-1.6727818453862 \end{cases}$$

**Detailed Steps on Solution **

## Solve the quadratic equation: $$x² - 33x - 58 = 0$$

Given $$a =1, b=-33, c=-58$$,

### 1. Use the quadratic root

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

$$\begin{aligned} \\x&=\dfrac{-b\pm\sqrt{b^2-4ac} }{2a}\\ & =\dfrac{-(-33)\pm\sqrt{(-33)^2-4\cdot 1\cdot (-58)}}{2 \cdot 1}\\ & =\dfrac{33\pm\sqrt{1321}}{2}\\ & =\dfrac{33}{2}\pm\dfrac{\sqrt{1321}}{2}\\ \end{aligned}$$

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

$$\begin{cases} x_1 =\dfrac{33}{2}+\dfrac{\sqrt{1321}}{2} \\ x_2=\dfrac{33}{2}-\dfrac{\sqrt{1321}}{2} \end{cases}$$

### 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,

$$x² - 33x - 58 = 0$$

Move the constant term $$-58$$ to the right hand side. Then its sign becomes postive.

$$x^2-33x=58$$

Add square of the half of $$-33$$, the coefficient of the linear term to both sides.

$$x^2-33x+\Big(\dfrac{33}{2}\Big)^2=58+\Big(\dfrac{33}{2}\Big)^2$$

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

$$\Big(x-\dfrac{33}{2}\Big)^2=\dfrac{1321}{4}$$

Taking square roots on both sides of above equation gives

$$\sqrt{\Big(x-\dfrac{33}{2}\Big)^2}=\pm\sqrt{\dfrac{1321}{4}}$$

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

$$x-\dfrac{33}{2}=\pm\dfrac{\sqrt{1321}}{2}$$

Move the constant $$-\dfrac{33}{2}$$ to the right hand side. Then we get,

$$x_1 = 17$$

$$x_2 = 16$$

### 3. The vertex of the function $$f(x) = x² - 33x - 58$$

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, $$a=1$$, $$b=-33$$ and $$c=-58$$. Since $$a >0$$, the curve of the function has the vertex at its lowest point. Substitute them to the vertex formula.

$$\begin{aligned} \\x&=\dfrac{-b}{2a}\\ & =\dfrac{-(-33)}{2\cdot 1}\\ & =\dfrac{33}{2}\\ \end{aligned}$$

$$\begin{aligned} \\y_{min}&=\dfrac{4ac-b^2}{4a}\\ & =\dfrac{4\cdot 1\cdot(-58)-(-33)^2}{4\cdot1^2}\\ & =-\dfrac{1321}{4}\\ \end{aligned}$$

So the coordinates for the vertex of the quadrautic function are $$\Big(\dfrac{33}{2},-\dfrac{1321}{4}\Big)$$

### 4. Graph for the function $$f(x) = x² - 33x - 58$$

Since the discriminat is greater than zero, the curve of the cubic function $$f(x) = x² - 33x - 58$$ has two intersection point with the x-axis