# How do you solve quadratic equation 4x^2+11x-20=0?

Nov 26, 2016

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

#### Explanation:

The first method to check for solving a quadratic equation is whether the expression factorises.

$4 {x}^{2} + 11 x - 20 = 0$

"Find factors of 4 and 20 which subtract to make 11"

Note that 11 is ODD, so the factors must combine to give one ODD and and one even number.

That immediately eliminates $2 \times 2$ and $10 \times 2$ as possible factors of 4 and 20
( because their multiples will always be even.)

When trying different combinations, remember not to have a common factor in any horizontal row.

Find factors and cross-multiply. Subtract the products to get 11.

" "ul(4" "20)
$\text{ "4" "5" } \rightarrow 1 \times 5 = 5$
$\text{ "1" "4" } \rightarrow 4 \times 4 = \underline{16}$
$\textcolor{w h i t e}{\times \times \times \times \times \times \times \times \times x} 11$ the difference is 11

We have the correct factors, now work with the signs.

MINUS $20$ means that the signs must be different.
PLUS $11$ means there must be more positives.
Fill in the correct signs, starting from $\textcolor{red}{+ 11}$

$\textcolor{red}{+ 11} \text{ } \rightarrow \textcolor{red}{+ 16} \mathmr{and} \textcolor{b l u e}{- 5}$

" "ul(4" "20)
$\text{ "4" "5" } \rightarrow 1 \times 5 = \textcolor{b l u e}{- 5}$
$\text{ "1" "4" } \rightarrow 4 \times 4 = \underline{\textcolor{red}{+ 16}}$
$\textcolor{w h i t e}{\times \times \times \times \times \times \times \times \times x} \textcolor{red}{+ 11}$

Now fill in the signs next to the correct factors:

" "ul(4" "20)
$\text{ "4" "color(blue)(-5)" } \rightarrow 1 \times \textcolor{b l u e}{- 5} = \textcolor{b l u e}{- 5}$
$\text{ "1" "color(red)(+4)" } \rightarrow 4 \times \textcolor{red}{+ 4} = \underline{\textcolor{red}{+ 16}}$
$\textcolor{w h i t e}{\times \times \times \times \times \times \times \times \times \times x} \textcolor{red}{+ 11}$

Now you have the factors: Top row is one bracket and bottom row is the other factor.

$4 {x}^{2} + 11 x - 20 = 0$
$\left(4 x - 5\right) \left(x + 4\right) = 0$

Letting each factor be equal to 0 gives the 2 solutions

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

Jan 12, 2017

$\frac{5}{4} \mathmr{and} - 4$
$y = 4 {x}^{2} + 11 x - 20 = 0$.
Transformed equation: $y ' = {x}^{2} + 11 x - 80 = 0$.
The 2 real roots of y are: $\frac{5}{4}$ and $- \frac{16}{4} = - 4$