1.what are the roots of the following equations? (i) 4/9p^2=2/3p (ii) 16m^2=4/25m?

Question

1.what are the roots of the following equations?
(i) 4/9p^2=2/3p (ii) 16m^2=4/25m
2.solve: 4z^2+4z+1=0
3. find x in terms of y
1/4x^2+xy+y^2=0
(show every step please)

Answer 1

(i) 4/9p²=2/3p

4/9p²-2/3p = 0

((2/3)p)*[(2/3)p-1] = 0

(2/3)p = 0 or [(2/3)p-1] = 0

p = 0 or (2/3)p = 1

p = 0 or 3/2
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2.solve: 4z²+4z+1=0

4z²+4z+1=0

4z²+2z + 2z+1=0

2z(2z + 1) +1(2z+1) = 0

(2z+1)(2z+1) = 0

(2z+1)² = 0

(2z+1) = 0

z = (-1/2)
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3. find x in terms of y
(1/4)x²+xy+y²=0

x²+4xy+4y²=0

x²+4xy+4y²=0

x²+2xy + 2xy+4y²=0

x(x+2y) + 2y(x+2y) = 0

(x+2y)(x+2y) = 0

(x+2y)²=0

(x+2y) = 0

x = -2y

Answer 2

(i) 4/9p^2=2/3p
4/9p^2-2/3p=0
p(4/9p-2/3)=0
p=0 or 4/9p-2/3=0
p=0 or 4/9p=2/3, p=3/2

p=0 or p=3/2

(ii) 4z^2+4z+1=0
(2z+1)*(2z+1)=0
z=-1/2

(iii) 1/4x^2+xy+y^2=0
y^2+xy+1/4x^2=0
(y+1/2x)*(y+1/2x)=0
y=-0,5x

Answer 3

1.4/9p^2=2/3p
=>2/3p=1 [dividing both sides by 2/3p]
=>3p=2
=>p=2/3
2.4z^2+4z+1=0
=>(2z)^2+2*2z*1+(1)^2=0
=>(2z+1)^2=0
=>2z+1=0
=>2z=-1 =>z=-1/2
3.Given exp.=(1/2x)^2+2*1/x*y+(y)^2=0
=>(1/2x+y)^2=0
=>1/2x+y=0
=1/2x=-y
=>x= -2y