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## $$x^2-17x-29=0$$

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

$$\Delta=405$$

$$\begin{cases} x_1=\dfrac{17}{2}+\dfrac{9}{2}\sqrt{5} \\ x_2=\dfrac{17}{2}-\dfrac{9}{2}\sqrt{5} \end{cases}$$

In decimal notation,

$$\begin{cases} x_1=18.562305898749 \\ x_2=-1.5623058987491 \end{cases}$$

Detailed Steps on Solution

## Solve the quadratic equation: $$x² - 17x - 29 = 0$$

Given $$a =1, b=-17, c=-29$$,

### 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{-(-17)\pm\sqrt{(-17)^2-4\cdot 1\cdot (-29)}}{2 \cdot 1}\\ & =\dfrac{17\pm9\sqrt{5}}{2}\\ & =\dfrac{17}{2}\pm\dfrac{9}{2}\sqrt{5}\\ \end{aligned}

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

That is,

$$\begin{cases} x_1 =\dfrac{17}{2}+\dfrac{9}{2}\sqrt{5} \\ x_2=\dfrac{17}{2}-\dfrac{9}{2}\sqrt{5} \end{cases}$$

### 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,

$$x² - 17x - 29 = 0$$

Move the constant term $$-29$$ to the right hand side. Then its sign becomes postive.

$$x^2-17x=29$$

Add square of the half of $$-17$$, the coefficient of the linear term to both sides.

$$x^2-17x+\Big(\dfrac{17}{2}\Big)^2=29+\Big(\dfrac{17}{2}\Big)^2$$

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

$$\Big(x-\dfrac{17}{2}\Big)^2=\dfrac{405}{4}$$

Taking square roots on both sides of above equation gives

$$\sqrt{\Big(x-\dfrac{17}{2}\Big)^2}=\pm\sqrt{\dfrac{405}{4}}$$

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

$$x-\dfrac{17}{2}=\pm\dfrac{9}{2}\sqrt{5}$$

Move the constant $$-\dfrac{17}{2}$$ to the right hand side. Then we get,

$$x_1 = 13$$

$$x_2 = 4$$

### 3. The vertex of the function $$f(x) = x² - 17x - 29$$

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, $$a=1$$, $$b=-17$$ and $$c=-29$$. 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{-(-17)}{2\cdot 1}\\ & =\dfrac{17}{2}\\ \end{aligned}

\begin{aligned} \\y_{min}&=\dfrac{4ac-b^2}{4a}\\ & =\dfrac{4\cdot 1\cdot(-29)-(-17)^2}{4\cdot1^2}\\ & =-\dfrac{405}{4}\\ \end{aligned}

So the coordinates for the vertex of the quadrautic function are $$\Big(\dfrac{17}{2},-\dfrac{405}{4}\Big)$$

### 4. Graph for the function $$f(x) = x² - 17x - 29$$

Since the discriminat is greater than zero, the curve of the cubic function $$f(x) = x² - 17x - 29$$ has two intersection point with the x-axis