# Which value of b would make 16x^2 -bx+25 a perfect square trinomial?

Jan 11, 2017

b = 40 and -40

#### Explanation:

General form of Perfect square trinomial is ${a}^{2} + 2 a b + {b}^{2}$

Therefore from
$16 {x}^{2} - b x + 25$

${a}^{2} = \sqrt{16 {x}^{2}} , {b}^{2} = 25$, then

$a = \pm 4 x , b = \pm 5$

take consideration a=4x and b=-5 (different sign), then
$- b x = 2 \left(4 x\right) \left(- 5\right)$
$- b x = - 40 x$
$b = 40$
The perfect square is ${\left(4 x - 5\right)}^{2} = 16 {x}^{2} - 40 x + 25$.

if we consider a=4x and b=5 (same sign), then
$- b x = 2 \left(4 x\right) \left(5\right)$
$- b x = 40 x$
$b = - 40$
The perfect square is ${\left(4 x + 5\right)}^{2} = 16 {x}^{2} + 40 x + 25$.

The first solution ${\left(4 x - 5\right)}^{2}$ is the best solution after comparing the expression given.
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