﻿ Solve 5x^2+35x-32=0 | Uniteasy.com

## $$5x^2+35x-32=0$$

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

$$\Delta=1865$$

$$\begin{cases} x_1=-\dfrac{7}{2}+\dfrac{\sqrt{1865}}{10} \\ x_2=-\dfrac{7}{2}-\dfrac{\sqrt{1865}}{10} \end{cases}$$

In decimal notation,

$$\begin{cases} x_1=0.81856457633784 \\ x_2=-7.8185645763378 \end{cases}$$

Detailed Steps on Solution

## Solve the quadratic equation: $$5x² + 35x - 32 = 0$$

Given $$a =5, b=35, c=-32$$,

### 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{-35\pm\sqrt{35^2-4\cdot 5\cdot (-32)}}{2 \cdot 5}\\ & =\dfrac{-35\pm\sqrt{1865}}{10}\\ & =-\dfrac{7}{2}\pm\dfrac{\sqrt{1865}}{10}\\ \end{aligned}

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

That is,

$$\begin{cases} x_1 =-\dfrac{7}{2}+\dfrac{\sqrt{1865}}{10} \\ x_2=-\dfrac{7}{2}-\dfrac{\sqrt{1865}}{10} \end{cases}$$

### 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,

$$5x² + 35x - 32 = 0$$

divide each term by $$5$$ to make the coefficient of the leading term $$1$$.

$$x^2+7x-\dfrac{32}{5}=0$$

Move the constant term $$-\dfrac{32}{5}$$ to the right hand side. Then its sign becomes postive.

$$x^2+7x=\dfrac{32}{5}$$

Add square of the half of $$7$$, the coefficient of the linear term to both sides.

$$x^2+7x+\Big(\dfrac{7}{2}\Big)^2=\dfrac{32}{5}+\Big(\dfrac{7}{2}\Big)^2$$

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

$$\Big(x+\dfrac{7}{2}\Big)^2=\dfrac{373}{20}$$

Taking square roots on both sides of above equation gives

$$\sqrt{\Big(x+\dfrac{7}{2}\Big)^2}=\pm\sqrt{\dfrac{373}{20}}$$

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

$$x+\dfrac{7}{2}=\pm\dfrac{1}{2}\sqrt{\dfrac{373}{5}}$$

Move the constant $$\dfrac{7}{2}$$ to the right hand side. Then we get,

$$x_1 = -3$$

$$x_2 = -4$$

### 3. The vertex of the function $$f(x) = 5x² + 35x - 32$$

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, $$a=5$$, $$b=35$$ and $$c=-32$$. 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{-35}{2\cdot 5}\\ & =-\dfrac{7}{2}\\ \end{aligned}

\begin{aligned} \\y_{min}&=\dfrac{4ac-b^2}{4a}\\ & =\dfrac{4\cdot 5\cdot(-32)-35^2}{4\cdot5^2}\\ & =-\dfrac{373}{4}\\ \end{aligned}

So the coordinates for the vertex of the quadrautic function are $$\Big(-\dfrac{7}{2},-\dfrac{373}{4}\Big)$$

### 4. Graph for the function $$f(x) = 5x² + 35x - 32$$

Since the discriminat is greater than zero, the curve of the cubic function $$f(x) = 5x² + 35x - 32$$ has two intersection point with the x-axis

$$5x^2+35x-31=0$$
$$5x^2-35x-31=0$$