Answer: There are no real roots.
Step-by-step explanation:
To find the roots of the function
f(x) = (2^x − 1) - (x2 + 2x − 3) with x ∈ R.
First open the bracket
2^x - 1 - x^2 - 2x + 3 = 0
Rearrange and collect the like terms
2x^2 - x^2 - 2x + 3 - 1= 0
X^2 - 2x + 2 = 0
Factorizing the above equation will be impossible, we can therefore find the root by using completing the square method or the quadratic formula.
X^2 - 2x = - 2
Half of coefficient of x is 1
X^2 - 2x + 1^2 = -2 + 1^2
( x - 1 )^2 = - 1
( x - 1 ) = +/- sqrt(-1)
X = -1 + sqrt (-1) or -1 - sqrt (-1)
The root of the function is therefore
X = -1 + sqrt (-1) or -1 - sqrt (-1)
Since b^2 - 4ac of the function is less than zero, we can therefore conclude that there is no real roots
Answer:
27 and 28
Step-by-step explanation:
Let a, ß be the roots of x² + 2ax + b =0.......[1]
⇒a + ß = -2a and aß = b
By hypothesis
la-ß | ≤ 2m
⇒ (a-ß)² ≤ 4m²
⇒ (a+ß)²-4aß < 4m²
⇒ 4a²-4b ≤ 4m²
⇒a² - b ≤ m²......(2)
And discriminant of (1) is >0
⇒4a²-4b> 0
⇒ b <a²...... (3)
From (2) and (3)
a² > b≥ (a² - m²)
b = [a² - m², a²)
Answer:
-a+-2b+4c
Step-by-step explanation:
Combine like terms
-4a+ 3a= -a
b+ -3b= -2b
3c+ c= 4c
