help me i need this today or i will get kick out (y)+9)(2y-3)=0?

Question

#1 (y)+9)(2y-3)=0 #2 m^2 +2m-15=0 #3 8r^2=24r #3 b^2-49=0 # 4 c(3c+16)=0 #5 z^4+8z^3 - 9z^2=0 #6 9y -7)(2y^2+7y-15)=0 please help they are going to kick me out of school please

why dont u help

Answer 1

Answer 1:
(y + 9)(2y - 3) = 0

Either y + 9 = 0 (or) 2y - 3 = 0
y = -9 (or) y = 3/2


Answer 2:
m^2 + 2m - 15 = 0
m^2 + 5m - 3m - 15 = 0
m(m + 5) - 3(m + 5) = 0
(m - 3)(m + 5) = 0

Either m - 3 = 0 (or) m + 5 = 0
m = 3 (or) m= -5


Answer 3:
8r^2 = 24r
r^2 = 3r
r^2 - 3r = 0
r(r - 3) = 0

Either r = 0 (or) r - 3 = 0
r = 0 (or) r = 3


Answer 4:
b^2 - 49 = 0
(b + 7)(b - 7) = 0
Either b + 7 = 0 (or) b - 7 = 0
b = -7 (or) 7


Answer 5:
c(3c + 16) = 0

Either c = 0 (or) 3c + 16 = 0
c = 0 (or) c = -16/3


Answer 6:
z^4 + 8z^3 - 9z^2 = 0
Even though this is biquadratic, we can solve it using quadratic equation knowledge
Factor out z^2

z^4 + 8z^3 - 9z^2 = 0
z^2(z^2 + 8z - 9) = 0

Either z^2 = 0 (or) z^2 + 8z - 9 = 0
Either z = 0 (or) z^2 + 8z - 9 = 0

We have one solution (z = 0) Solve z^2 + 8z - 9 = 0 to get the other two solutions.

z^2 + 8z - 9 = 0
z^2 + 9z - z - 9 = 0
z(z + 9) - 1(z + 9) = 0
(z + 9)(z - 1) = 0

z = -9 (or) z = 1
The three solutions are z = 0, 1, -9


Answer 7:
(9y - 7)(2y^2 + 7y - 15) = 0
Either 9y - 7 = 0 (or) 2y^2 + 7y - 15 = 0
y = 7/9 (or) 2y^2 + 7y - 15 = 0

2y^2 + 7y - 15 = 0
2y^2 + 10y - 3y - 15 = 0
2y(y + 5) - 3(y + 5) = 0
(2y - 3)(y + 5) = 0
y = 3/2 (or) y = -5

The solutions are y = 7/9, 3/2, -5

Answer 2

Here is your answer,

1. ( y+9)(2y-3) =0 so, y+9= 0 which gives y= -9
also, 2y-3 = 0 which gives 2y=3 or y=3/2

2. m^2 + 2m -15 = 0
or, m^2 +5m -3m -15 = 0
or, m(m +5) + 3( m+5)= 0
or, ( m+5)(m+3 ) = 0 so , m+5 = 0 or, m= -5
also, m+3= 0 or, m=-3

3. 8r^2= 24 r
or, 8r^2 -24 r = 0
or, 8r(r-3)= 0 so , 8 r = 0 or r= 0
also r-3 = 0 or r=3

4. c(3c + 16) = 0
so , c= 0
also , 3c+ 16 = 0
or, 3c = -16
or, c = - 16/3

5. z^4 + 8z^3 - 9z^2 = 0
z^2 ( z^2 + 8 z - 9) = 0 so, z^2 = 0 or, z = 0
also, z^2 + 8z -9 = 0
z^2 +9z_-1z -9 = 0
z( z+9) - 1(z+9) = 0
( z+9)( z-1)= 0
so , z--1= 0 or z = 1
also , z+9 = 0 or, z = -9

6. ( 9y- 7)(2y^2+7y-15 ) = 0

so, 9y- 7 = 0 or, 9y = 7 or, y = 7/9
also, 2y^2 +7y -15 = 0
2y^2 +10 y - 3y - 15 = 0
2y(y+5) - 3(y+5)= 0
( 2y-3)(y+5) = 0 so, 2y -3 = 0 or, 2y = 3
or, y = 3/2

also y+5 = 0 or, y = -5

Answer 3

#1 (y+9)(2y-3)=0
Either (y+9)=0 --> y = -9
or 2y-3=0 --> y=3/2
Ans. y= -9, 3/2

#2 m^2+2m-15=0
m^2 +5m -3m -15=0
m(m+5)-3(m+5)=0
(m+5)(m-3)=0
m+5=0 --> m= -5
m-3=0 --> m=3
Ans. m= -5, 3

#3 8r^2=24r
8r^2-24r=0
8r(r-3)=0
r=0
r-3=0 --> r=3
Ans. r=0, 3

#3 b^2-49=0
b^2-7^2=0
(b+7)(b-7)=0
b+7=0 --. b= -7
b-7=0 --> b= 7
Ans. b=7, -7

#4 c(3c+16)=0
Either c=0
or 3c+16=0 -> 3c=-16 -> c=-16/3 -> c= -5 1/3
Ans. c=0, -5 1/3

#5 z^4+8z^3-9z^2=0
z^2(z^2+8z-9)=0
z^2(z^2+9z-z-9)=0
z^2[z(z+9)-(z+9)]=0
z^2[(z+9)(z-1)]=0
Either z^2=0 or z+9=0 or z-1=0
Ans.z=0, -9, 1

#6 (9y-7)(2y^2+7y-15)=0
(9y-7)(2y^2+10y-3y-15)=0
(9y-7)[(2y(y+5)-3(y+5)]=0
(9y-7)[(y+5)(2y-3)]=0
Either 9y-7=0 --> y=7/9
or (y+5)=0 --> y= -5
or (2y-3)=0 --> y=3/2 or 1 1/2
Ans. y= 7/9, -5, 1 1/2

Best of luck

Answer 4

1. What are the instructions? If you're asked to solve the equation, then set each factor equal to zero & isolate the variable.
2. Factor: you need two factors of -15 that add up to 2.
-15 = (-3)(5)
-15 = (3)(-5)
Which pair adds up to 2?

Someone may provide all the answers, but that won't help you.

Answer 5

maybe you should learn how to solve quadratic equations so you can stay in school. NO WONDER they are trying to get rid of you - you obviously have no interest in LEARNING!

come on! Rally! buckle down and LEARN!