# How do you solve 4x^2+17x+15=0?

Jun 11, 2018

$x = - 3$ or $x = - \frac{5}{4}$

#### Explanation:

$4 {x}^{2} + 17 x + 15 = 0$

Do

$4 {x}^{2} + 12 x + 5 x + 15 = 0$

Then

$4 x \left(x + 3\right) + 5 \left(x + 3\right) = 0$

$\left(4 x + 5\right) \left(x + 3\right) = 0$

So

$x = - 3 \text{ }$ or $\text{ } x = - \frac{5}{4}$

Jun 11, 2018

$- 3 , \mathmr{and} - \frac{5}{4}$

#### Explanation:

Method 1. Use the improved quadratic formula
$y = 4 {x}^{2} + 17 x + 15 = 0$.
$D = {d}^{2} = {b}^{2} - 4 a c = 289 - 240 = 49$ --> $d = \pm 7$
There are 2 real roots:
$x = - \frac{b}{2 a} \pm \frac{d}{2 a} = - \frac{17}{8} \pm \frac{7}{8}$
$x 1 = - \frac{24}{8} = - 3$
$x 2 = - \frac{10}{8} = - \frac{5}{4}$
Method 2. Use the new Transforming Method (Socratic, Google Search)
Transformed equation: $y ' = {x}^{2} + 17 x + 60$.
Proceeding. Find the 2 real roots of y', then, divide them by a = 4.
Find 2 real roots, bot negative (Rule of signs), knowing the sum (-b = -17), and the product (ac = 60). They are - 5 and - 12.
The 2 real roots of y are:
$x 1 = - \frac{5}{a} = - \frac{5}{4}$, and $x 2 = - \frac{12}{a} = - \frac{12}{4} = - 3$

Jun 11, 2018

$x - 3 \mathmr{and} x = - \frac{5}{4}$

#### Explanation:

Find factors of $4 \mathmr{and} 15$ whose products add to $17$

$\text{ } 4 \mathmr{and} 15$
$\text{ "darr" } \downarrow$
$\text{ "4" "5" } \rightarrow 1 \times 5 = 5$
$\text{ "1" "3" } \rightarrow 4 \times 3 = \underline{12}$
$\textcolor{w h i t e}{\times \times \times \times \times \times \times \times x} 17$

The signs are all positive.

$\left(4 x + 5\right) \left(x + 3\right) = 0$

Either factor can be equal to $0$

If $4 x + 5 = 0 \text{ } \rightarrow x = - \frac{5}{4}$

If $x + 3 = 0 \text{ } \rightarrow x = - 3$