How do you find the roots, real and imaginary, of $h=--t^2+16t -4$ using the quadratic formula?

Question

1406 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-06-05T18:00:10

$color(blue)(h=-8+2sqrt17 or h=-8-2sqrt17$

$h=- -t^2+16t-4$

$h=-(-t^2)+16t-4$

$h=t^2+16t-4$

$x=(-b+-sqrt(b^2-4ac))/(2a)$

$color(blue)(a^2+bx+c$

$color(blue)(a=1,b=16,c=-4$

$h=(-(16)+-sqrt((16)^2-4(1)(-4)))/(2(1))$

$(-16+-sqrt(256+16))/2$

$(-16+-sqrt(272))/2$

$(-16+-sqrt(2*2*2*2*17))/2$

$color(blue)(sqrt2 xx sqrt2=2$

$(-16+-4sqrt17)/2$

$(cancel4^color(blue)2(-4+sqrt17))/cancelcolor(blue)2^color(blue)1 or (cancel4^color(blue)2(-4-sqrt17))/cancel2^color(blue)1$

$2(-4+sqrt17) or 2(-4-sqrt17))$

$color(blue)(-8+2sqrt17 or -8-2sqrt17$

How do you find the roots, real and imaginary, of h=--t^2+16t -4 h=--t^2+16t -4 using the quadratic formula?