Math Problems?

Question

Can someone solve these math problems for me. Thanks!

r^2 = 5 - 4r

p^2 - p = 0

4t^2 - 9 = 0

Also, please show me how you got your answer to each problem. Thanks again!

Answer 1

1)r^2=5-4r
=>r^2+4r-5=0
=>r^2+5r-r-5=0
=>r(r+5)-1(r+5)=0
=>(r+5)(r-1)=0
Therefore either r+5=0 or r-1=0
If r=5=0,then r= -5
If r-1=0.then r=1
r= -5 or 1 ans
2)p^2-p=0
=>p^2=p
=>p=1 [dividing both sides by p]
3)4t^2-9=0
=>4t^2=9
=>t^2=9/4
t=+- sqrt(9/4)[square-rooting both sides]
=+-3/2=+-1 1/2 ans

Answer 2

If you need to show your work:

r^2 + 4r - 5 =0
(r+5) (r-1)=0
r= -5, 1

p^2 -p=0
p(p-1)=0
p=0, 1

4t^2-9=0
take square root of both sides
2t-3=0
t=3/2

Answer 3

These are quadratics so transpose to the familiar form:

r^2=5 - 4r
r^2+4r-5=0

now solve by inspection (remember FOIL)

(r-1)(r+5)=0
so
r-1=0
r=1

and

r+5=0
r=-5

Try the others yourself

Answer 4

r² + 4r - 5 = 0
(r + 5).(r - 1) = 0
r = - 5, r = 1

p.(p - 1) = 0
p = 0, p = 1

(2t - 3).(2t + 3) = 0
t = 3 / 2, t = - 3 / 2

Answer 5

r = -5 or 1

p = 0 or 1

t = 1.5 or -1.5

Answer 6

r = -5 or 1
p = 0 or 1
t = 1.5 or -1.5