7x³ = 28x is our equation. We want its solutions.
When you have x and different powers, set the whole thing equal to zero.
7x³ = 28x
7x³ - 28x = 0
Now notice there's a common x in both terms. Let's factor it out.
x (7x² - 28) = 0
As 7 is a factor of 7 and 28, it too can be factored out.
x (7) (x² - 4) = 0
We can further factor x² - 4. We want a pair of numbers that multiply to 4 and whose sum is zero. The pairs are 1 and 4, 2 and 2. If we add 2 and -2 we get zero.
x (7) (x - 2) (x + 2) = 0
Now we use the Zero Product Property - if some product multiplies to zero, so do its pieces.
x = 0 -----> so x = 0
7 = 0 -----> no solution
x - 2 = 0 ----> so x = 2 after adding 2 to both sides
x + 2 = 0 ---> so = x - 2 after subtracting 2 to both sides
Thus the solutions are x = 0, x = 2, x = -2.
M’ (-4, 7)
N’ (0, 5)
P’ (2, 0)
Q’ (-2, 1)
you basically change the x and y values by the value it says so you'd subtract 4 from x (the first number) and add 1 to y (the second number)
Answer:
4y = x - 17
Step-by-step explanation:
First off, in the equation given : 2x + 5y = 6 , make y the subject of the formula which will give y = 
+ 
which make the gradient, m, the coefficient of x = 
. Since the line we're looking for is parallel to the one we were given, their respective gradients would be the same. (If they were perpendicular, the gradient of the new line would be a negative inverse of the given line)
Then you proceed to use the one-point formula : y - y₀ = m(x - x₀), where y₀ = 3 and x₀ = 5 from the point it goes through (5, 3)
y - 3 = 
(x - 5); y - 3 = 
- 
; y = - +3
y = - 
; y = 
; 4y = x - 17
Answer:
<h2>8 or -8</h2>
Step-by-step explanation:
The absolute value of number a:
|a| = a for a ≥ 0
|a| = -a for a < 0
|a| = 8 ⇔ a = 8 or a = -8
