# Solve the quadraic equation:

## $$x^2-10x-27=0 $$

**Quick Answer**

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

$$ \Delta=208$$

$$\begin{cases} x_1=5+2\sqrt{13} \\ x_2=5-2\sqrt{13} \end{cases}$$

In decimal notation,

$$\begin{cases} x_1=12.211102550928 \\ x_2=-2.211102550928 \end{cases}$$

**Detailed Steps on Solution **

## Solve the quadratic equation: $$x² - 10x - 27 = 0$$

Given $$a =1, b=-10, c=-27$$,

### 1. Use the quadratic root

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

$$\begin{aligned} \\x&=\dfrac{-b\pm\sqrt{b^2-4ac} }{2a}\\ & =\dfrac{-(-10)\pm\sqrt{(-10)^2-4\cdot 1\cdot (-27)}}{2 \cdot 1}\\ & =\dfrac{10\pm4\sqrt{13}}{2}\\ & =5\pm2\sqrt{13}\\ \end{aligned}$$

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

$$\begin{cases} x_1 =5+2\sqrt{13} \\ x_2=5-2\sqrt{13} \end{cases}$$

### 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,

$$x² - 10x - 27 = 0$$

Move the constant term $$-27$$ to the right hand side. Then its sign becomes postive.

$$x^2-10x=27$$

Add square of the half of $$-10$$, the coefficient of the linear term to both sides.

$$x^2-10x+\Big(5\Big)^2=27+\Big(5\Big)^2$$

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

$$\Big(x-5\Big)^2=52$$

Taking square roots on both sides of above equation gives

$$\sqrt{\Big(x-5\Big)^2}=\pm\sqrt{52}$$

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

$$x-5=\pm2\sqrt{13}$$

Move the constant $$-5$$ to the right hand side. Then we get,

$$x_1 = 7$$

$$x_2 = 3$$

### 3. The vertex of the function $$f(x) = x² - 10x - 27$$

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, $$a=1$$, $$b=-10$$ and $$c=-27$$. Since $$a >0$$, the curve of the function has the vertex at its lowest point. Substitute them to the vertex formula.

$$\begin{aligned} \\x&=\dfrac{-b}{2a}\\ & =\dfrac{-(-10)}{2\cdot 1}\\ & =5\\ \end{aligned}$$

$$\begin{aligned} \\y_{min}&=\dfrac{4ac-b^2}{4a}\\ & =\dfrac{4\cdot 1\cdot(-27)-(-10)^2}{4\cdot1^2}\\ & =-52\\ \end{aligned}$$

So the coordinates for the vertex of the quadrautic function are $$\Big(5,-52\Big)$$

### 4. Graph for the function $$f(x) = x² - 10x - 27$$

Since the discriminat is greater than zero, the curve of the cubic function $$f(x) = x² - 10x - 27$$ has two intersection point with the x-axis