# How do you solve 6x^2+5x=1?

Apr 29, 2017

Re-arranging we get:

$6 {x}^{2} + 5 x - 1 = 0$

We can write this as:

$6 {x}^{2} + 6 x - x - 1 = 0$

i.e. $6 x \left(x + 1\right) - \left(x + 1\right) = 0$

=> $\left(6 x - 1\right) \left(x + 1\right) = 0$

so, $x = \frac{1}{6} \mathmr{and} x = - 1$

:)>

Apr 29, 2017

$x = - 1 \text{ or } x = \frac{1}{6}$

#### Explanation:

$\text{subtract 1 from both sides and equate to zero}$

$6 {x}^{2} + 5 x - 1 = 1 - 1$

$\Rightarrow 6 {x}^{2} + 5 x - 1 = 0$

$\text{using the a-c method of factorising}$

$\text{require the factors of the product " -6" which sum to +5}$

$\text{the factors are } + 6 , - 1$

$\Rightarrow 6 {x}^{2} + 6 x - x - 1 = 0 \leftarrow \text{ split 5x}$

$\Rightarrow 6 x \left(x + 1\right) - 1 \left(x + 1\right) = 0 \leftarrow \text{ factor pairs}$

$\left(x + 1\right) \left(6 x - 1\right) = 0 \leftarrow \text{ factor out (x+1)}$

$\left(x + 1\right) = 0 \Rightarrow x = - 1$

$\left(6 x - 1\right) = 0 \Rightarrow x = \frac{1}{6}$

$\Rightarrow x = - 1 \text{ or " x=1/6" are the solutions}$