100t -16t2 = 0 t4(25-4t) = 0 t = 0 or t = 25/4
Unfortunately you cannot solve this problem since the domain of x (b2-4ac) = 4-4*2*1 = -4Since the root of a negative number does not exist, this problem is nog solvable without the use of imaginary numbers.
Suppose x3-4x = 0. To solve, factor: x3-4x = x(x2-4) = x(x+2)(x-2) = 0 Now, a product equals 0 if and only one or more of the factors equals 0, so set each factor to 0 and solve. The roots are 0,-2 and +2.
(3x+4)(3x-4)=0 x=±4/3
Subtract x so -4 = 2x/ x= -2
y = 3, 6
100t -16t2 = 0 t4(25-4t) = 0 t = 0 or t = 25/4
y=4
9v2 - 4 = 0 Add 4 to both sides 9v2 = 4 Take square roots of both sides: ±3v = ±2 so that v = ±2/3
Unfortunately you cannot solve this problem since the domain of x (b2-4ac) = 4-4*2*1 = -4Since the root of a negative number does not exist, this problem is nog solvable without the use of imaginary numbers.
Suppose x3-4x = 0. To solve, factor: x3-4x = x(x2-4) = x(x+2)(x-2) = 0 Now, a product equals 0 if and only one or more of the factors equals 0, so set each factor to 0 and solve. The roots are 0,-2 and +2.
(3x+4)(3x-4)=0 x=±4/3
-4x-2+2=-2+2. -4x/-4=0/-4 x=0
Subtract x so -4 = 2x/ x= -2
x2+1/4x = 0 x(x+1/4) = 0 Therefore: x = 0 or x = -1/4
(4 ± i2) where i2 = -1
(x - 16)(x + 4) = 0 so x = -4 or 16