Answer:
b) 1
Step-by-step explanation:
First, bring the -3 to the right side.
x²+4x + 4
now use b²-4ac.
4²-4(1)(4)
16-16
0
If discriminant < 0 there are 2 imaginary solutions
if discriminant > 0 there are 2 real solutions
if discriminant = 0 there is 1 real solution
The discriminant here is 0 so it has one real solution
Assuming the function is
f(x) = x^4 + 3x^3-28x^2
Factorising, we get
f(x) = x^2(x+7)(x-4)
Therefore, there is a double root at x = 0 and singular roots at x = -7 and 4
double roots tend to bounce, kind of like the standard quadratic y = x^2
singular roots tend to cross, kind of like a straight line on a graph, y = mx + b
Therefore, it bounces at x = 0, and crosses at x = -7 and 4
In general, a root raised to an even power bounces, and a root raised to an odd power crosses.
Answer:
five more than twice a number 2n + 5
the product of a number and 6 6n
seven divided by twice a number 7 ÷ 2n or
three times a number decreased by 11 3n - 11
Step-by-step explanation: