If P is a root of the equation X3 +10X2 + 16X, than which of the following equations have also the root P ?
Answer(s): B
The original equation is X3 +10X2 + 16X, it can be written as X(X + 8)(X + 2). The roots are (-8),0 and (-2). We are looking for an equation that has one of the same roots.Answer B: X + 8 This equation has the root (-8).All the other answers have different roots than the original equation.
If X is a root of the equation a3 +8a2 – 20a, than which of the following equations Don’t have the root X as one of their roots?
Answer(s): E
The original equation is a3 +8a2 – 20a, it can be written as a(a – 2)(X + 10). The roots are 2,0 and (-10). We are looking for an equation that has none of the same roots.Answer E: X2 – 10X +16 = (X + 2)(X + 8). This equation has none of the original roots. All the other answers have one or more of the same original roots.
(x, y) are the coordinates of the intersection of the following lines:(3x – 2y = 8) and (3y + x = 10). What is the value of (x/y)?
There is no need to draw the lines. There are two equations with two variable that you have to solve. Take the second equation and multiply it by (-3) to get: -9y –3x = -30 add this equation to the first and You’ll get: -11y = -22 y=2 and x=4. (x/y) is 2.
A(a, b) is the coordinates of the intersection between the lines:(x + y –1 = 0) and (4x – 2y = 5). What is the shortest distance between A(a, b) and the coordinate B(25/6, 23/6)?
There is no need to draw the two lines. Multiply equation (1) by 2 and then add the equations to get: 6x = 7 . x = 7/6, y = -1/6.Draw a rectangular axis system and mark the point A and B.Complete the two points to a triangle so one of sides is 3 and the other is 4, the hypotenuse, which is also the requested length is 5.
P(x, y) is the intersection point between the circle (x2 + y2 = 4) and the line (y = x +2). Which of the following can be the point P?
Answer(s): D
First, draw the circle and the line. The circle is centered at (0, 0) with a radius of 2.You can see that the line and the circle intersect at two points: (-2, 0) and (0, 2). Another way is to insert y = x +2 into the equation of the circle and solve it.
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