***Note these are separate problems.
x(x-32)=0
4b(b+4)=0
(y-3)(y+2)=0
2006-12-23 09:28:13 · 6 answers · asked by baseballman1243 1 in Science & Mathematics ➔ Mathematics
x(x-32)=0
Well, if either x or (x-32) is equal to zero, then the equation is true. x is 0 if x=0 and (x-32) is zero if x=32. Answers: 0,32.
4b(b+4)=0
By the same logic, this is true if b=0 or b=-4
(y-3)(y+2)=0
y=3,-2
These problems (quadratic equations) rely on the zero property of multiplication.
2006-12-23 09:32:13 · answer #1 · answered by knock knock 3 · 0⤊ 0⤋
To make the left hand side = 0, then
x=0 or x =32
b=0 or b= -4
y=3 or y= -2
2006-12-23 09:32:47 · answer #2 · answered by ironduke8159 7 · 0⤊ 0⤋
you need to make one polynomial/monomial = 0
x(x-32)=0
x=0 or x=32
4b(b+4)=0
b=0 or b=4
(y-3)(y+2)=0
y=3 pr y= -2
2006-12-23 09:36:38 · answer #3 · answered by jekelly90 2 · 0⤊ 0⤋
1. First: solve for the "x" variables > set them to equal zero:
x = 0
x - 32 = 0
x - 32 + 32 = 0 + 32
x = 32
*Follow the same format for the other two.
2006-12-23 11:17:53 · answer #4 · answered by ♪♥Annie♥♪ 6 · 0⤊ 0⤋
x(x-32)=0
x=0, 32
4b(b+4)=0
b=0, -4
(y-3)(y+2)=0
y=3, -2
find the zeros thats all you had to do
2006-12-23 10:09:18 · answer #5 · answered by drunken_warrior 1 · 0⤊ 0⤋
x(x-32)=0
x=0
x-32=0
x=32
x=0, 32
4b(b+4)=0
4b=0
b=0
b+4=0
b=-4
b=0, -4
(y-3)(y+2)=0
y-3=0
y=3
y+2=0
y=-2
y=3, -2
2006-12-23 13:05:50 · answer #6 · answered by mu_do_in 3 · 0⤊ 0⤋