1.) In solving the equation (x + 1)(x - 2) = 4, Eric stated that the solution would be x + 1 = 4 => x = 3 or (x - 2) = 4 => x = 6. However, at least one of these solutions fails to work when substituted back into the original equation. Why is that? Please help Eric to understand better; solve the problem yourself, and explain your reasoning.

2.) If a stone is tossed from the top of a 330 meter building, the height of the stone as a function of time is given by h(t) = -9.8t2 - 10t + 330, where t is in seconds, and height is in meters. After how many seconds will the stone hit the ground? Round to the nearest hundredth's place; include units in your answer.

3.) Use the discriminant to determine whether the following equations have solutions that are: two different rational solutions; two different irrational solutions; exactly one rational solution; or two different imaginary solutions.
3x^2 + 6x - 5 = 0

Solution Preview

1. Eric should be making use of the Zero Product Theorem, but for that to work the product of his two factors must be 0. Here is the correct solution.

...solve for y: 0 + 4 y = −2 −2 y= 4 1 y=− 2 To solve for x, we will substitute y in either of the simplified equations. We choose equation 1: 1 2x ...

... system of graphing 1. y= -2/3x 2X + 3Y=9 Solve each system by ... Determine whether the equations are independent, dependent, or inconsistent 2. x= y + 3 ...

... y= -x/3+1. 6) Yes, the order pair is the solution as when we plug the pair to both equations, they both work. For example, for the first equation, plug (-4, 2/3 ...

... Check your solution by substituting your solution in the original equation. A: (x+2)(x-2)=x2-4 and 15+(x+1)(x-7)=15+x2-6x-7=8+x2-6x So x2-4=8+x2-6x. ...

... 5 = 0 f) s2 - 4s + 4 = 0 g) 5/6x2 - 7x - 6/5 = 0 h) 7a2 + 8a + 2 = 0 2. If x = 1 and x = -8, then form a quadratic equation. 3. What type of solution do you ...

... 0) = 0 and ∂t boundary conditions u (0, y, t ) = u (1, y, t ) = u ( x,0, t ) = u ( x,1, t ) = 0. 2. Solve the two-dimensional wave equation for a ...

... Case 1: a = k 2 > 0. Here the solution to the equation is: x (x ) = A e kx + Be - kx. Applying the boundary conditions: x (0) = A + B = 0 Þ A=-B. ...