# How do you solve (5x-1)^2=16?

Aug 25, 2016

$x = - \frac{3}{5}$ and $x = 1$

#### Explanation:

First you expand ${\left(5 x - 1\right)}^{2}$ so $\left(5 x - 1\right) \cdot \left(5 x - 1\right) = 25 {x}^{2} - 10 x + 1$
That leaves us with:
$25 {x}^{2} - 10 x + 1 = 16$
Now we subtract 16 from both sides:
$25 {x}^{2} - 10 x - 15 = 0$
Then all we have to do is plug a,b and c from $a {x}^{2} + b x + c$ into the quadratic formula $x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$
$a = 25$
$b = - 10$
$c = - 15$

$x = \frac{- \left(- 10\right) \pm \sqrt{{\left(- 10\right)}^{2} - 4 \cdot 25 \cdot \left(- 15\right)}}{2 \cdot 25}$

$x = \frac{10 \pm \sqrt{100 - \left(- 1500\right)}}{50}$

$x = \frac{10 \pm \sqrt{1600}}{50}$

$x = \frac{10 \pm 40}{50}$

So $x = \frac{50}{50} = 1$ and $x = - \frac{30}{50} = - \frac{3}{5}$

Take the square root of both sides and solve to get to $x = 1$

#### Explanation:

${\left(5 x - 1\right)}^{2} = 16$

The first problem we have is that everything on the left is squared. We could multiply it out and end up with $25 {x}^{2} - 10 x + 1$, but since the right side is a perfect square, let's instead take the square root of both sides:

$\sqrt{{\left(5 x - 1\right)}^{2}} = \sqrt{16}$

$5 x - 1 = 4$

And now let's solve:

$5 x = 5$

$x = 1$

And we can check the answer:

${\left(5 x - 1\right)}^{2} = 16$

${\left(5 \left(1\right) - 1\right)}^{2} = 16$

${\left(5 - 1\right)}^{2} = 16$

${\left(4\right)}^{2} = 16$

$16 = 16$