How do you find the roots, real and imaginary, of $h = - 16 t^{2} + 64 t$ $h=-16t^2+64t$ using the quadratic formula?

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Answer 1 · 2018-04-01T21:55:35

0, 4

The quadratic formula states
$x = \frac{- b \pm \sqrt b^{2} - 4 a c}{2 a}$ $x=(-b±√b^2-4ac)/(2a)$

In this case,
$a = - 16$ $a=-16$
$b = 64$ $b=64$
$c = 0$ $c=0$

Plug in the numbers.

To make it simpler, let's only substitute $a$ $a$ first.

$x = \frac{- b \pm \sqrt b^{2} - (- 64 c)}{- 32}$ $x=(-b±√b^2-(-64c))/-32$

Substitute $b$ $b$ .

$x = \frac{- 64 \pm \sqrt 64^{2} + 64 c}{- 32}$ $x=(-64±√64^2+64c)/-32$

Finally, substitute $c$ $c$ .

$x = \frac{- 64 \pm \sqrt 64^{2}}{- 32}$ $x=(-64±√64^2)/-32$

Simplify to get
$x = \frac{- 64 \pm 64}{- 32}$ $x=(-64±64)/-32$

If $x = \frac{- 64 + 64}{- 32}$ $x=(-64+64)/-32$ , then $x = 0$ $x=0$ because $\frac{0}{- 32} = 0$ $0/-32 = 0$ .

If $x = \frac{- 64 - 64}{- 32}$ $x=(-64-64)/-32$ , then $x = 4$ $x=4$ because $- \frac{128}{- 32} = 4$ $-128/-32=4$ .

The roots of $h = - 16 t^{2} + 64 t$ $h=-16t^2+64t$ are 0, 4.

Yay!

How do you find the roots, real and imaginary, of h=−16t2+64t h=-16t^2+64t using the quadratic formula?