| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 2.91 |
| Score | 0% | 58% |
Solve for x:
x2 - 8x - 4 = -x + 4
| 5 or 4 | |
| -4 or -7 | |
| -1 or 8 | |
| 6 or -1 |
The first step to solve a quadratic expression that's not set to zero is to solve the equation so that it is set to zero:
x2 - 8x - 4 = -x + 4
x2 - 8x - 4 - 4 = -x
x2 - 8x + x - 8 = 0
x2 - 7x - 8 = 0
Next, factor the quadratic equation:
x2 - 7x - 8 = 0
(x + 1)(x - 8) = 0
For this expression to be true, the left side of the expression must equal zero. Therefore, either (x + 1) or (x - 8) must equal zero:
If (x + 1) = 0, x must equal -1
If (x - 8) = 0, x must equal 8
So the solution is that x = -1 or 8
Breaking apart a quadratic expression into a pair of binomials is called:
deconstructing |
|
factoring |
|
squaring |
|
normalizing |
To factor a quadratic expression, apply the FOIL (First, Outside, Inside, Last) method in reverse.
If a = c = 3, b = d = 6, and the blue angle = 69°, what is the area of this parallelogram?
| 21 | |
| 27 | |
| 3 | |
| 18 |
The area of a parallelogram is equal to its length x width:
a = l x w
a = a x b
a = 3 x 6
a = 18
Solve for y:
y2 - 10y + 9 = 0
| 1 or -6 | |
| 1 or 9 | |
| 6 or 6 | |
| 8 or -2 |
The first step to solve a quadratic equation that's set to zero is to factor the quadratic equation:
y2 - 10y + 9 = 0
(y - 1)(y - 9) = 0
For this expression to be true, the left side of the expression must equal zero. Therefore, either (y - 1) or (y - 9) must equal zero:
If (y - 1) = 0, y must equal 1
If (y - 9) = 0, y must equal 9
So the solution is that y = 1 or 9
Find the value of a:
-7a + y = -6
-7a + 4y = -8
| -1 | |
| \(\frac{16}{21}\) | |
| -\(\frac{5}{7}\) | |
| 8\(\frac{1}{5}\) |
You need to find the value of a so solve the first equation in terms of y:
-7a + y = -6
y = -6 + 7a
then substitute the result (-6 - -7a) into the second equation:
-7a + 4(-6 + 7a) = -8
-7a + (4 x -6) + (4 x 7a) = -8
-7a - 24 + 28a = -8
-7a + 28a = -8 + 24
21a = 16
a = \( \frac{16}{21} \)
a = \(\frac{16}{21}\)