Answer:
(a) [tex]\sf -16r^2+20r-4=0[/tex]
(b) r = 1, r = 1/4
Step-by-step explanation:
Given equation: [tex]\sf h=-16r^2+20r-4[/tex]
where:
Question (a)
The ball will hit the ground when h = 0
[tex]\sf \implies-16r^2+20r-4=0[/tex]
Question (b)
To factor [tex]\sf -16r^2+20r-4=0[/tex]
Divide both sides by 4:
[tex]\sf \implies-4r^2+5r-1=0[/tex]
Swap signs:
[tex]\sf \implies 4r^2-5r+1=0[/tex]
Split middle term:
[tex]\sf \implies 4r^2-4r-r+1=0[/tex]
Factorize the first two terms and the last two terms separately:
[tex]\sf \implies 4r(r-1)-(r-1)=0[/tex]
Factor out common term (r - 1):
[tex]\sf \implies (4r-1)(r-1)=0[/tex]
Therefore:
[tex]\sf \implies (4r-1)=0 \implies r=\dfrac14[/tex]
[tex]\sf \implies (r-1)=0 \implies r=1[/tex]