# Given the quadratic function f(x) = x^2 - 12x + 36 how do you find a value of x such that f(x) = 25?

Jun 20, 2016

$\therefore x = 11 , x = 1.$

#### Explanation:

$f \left(x\right) = 25$
$\therefore {x}^{2} - 12 x + 36 = 25.$
$\therefore {x}^{2} - 12 x + 36 - 25 = 0.$
$\therefore {x}^{2} - 12 x + 11 = 0.$
$\therefore {x}^{2} - 11 x - 1 x + 11 = 0.$.........$\left[11 \times 1 = 11 , 11 + 1 = 12\right]$
$\therefore x \left(x - 11\right) - 1 \left(x - 11\right) = 0.$
$\therefore \left(x - 11\right) \left(x - 1\right) = 0.$
$\therefore x = 11 , x = 1.$

Alternatively, we see that $f \left(x\right) = {\left(x - 6\right)}^{2.}$
Hence, $f \left(x\right) = 25$
$\Rightarrow {\left(x - 6\right)}^{2} = 25 = {5}^{2.}$
$\Rightarrow \left(x - 6\right) = \pm 5$
$\Rightarrow x = 6 \pm 5$
$\Rightarrow x = 11 , \mathmr{and} , x = 1 ,$ as before!