Answer:
Roots are not real
Step-by-step explanation:
To prove : The roots of x^2 +(1-k)x+k-3=0x
2
+(1−k)x+k−3=0 are real for all real values of k ?
Solution :
The roots are real when discriminant is greater than equal to zero.
i.e. b^2-4ac\geq 0b
2
−4ac≥0
The quadratic equation x^2 +(1-k)x+k-3=0x
2
+(1−k)x+k−3=0
Here, a=1, b=1-k and c=k-3
Substitute the values,
We find the discriminant,
D=(1-k)^2-4(1)(k-3)D=(1−k)
2
−4(1)(k−3)
D=1+k^2-2k-4k+12D=1+k
2
−2k−4k+12
D=k^2-6k+13D=k
2
−6k+13
D=(k-(3+2i))(k+(3+2i))D=(k−(3+2i))(k+(3+2i))
For roots to be real, D ≥ 0
But the roots are imaginary therefore the roots of the given equation are not real for any value of k.
Answer:
y≤
Step-by-step explanation:
Since the graph is facing down, We know the equation is y<mx+b
Since slope is rise/run the slope is 1/3.
Since the y intercept (The point where the line hits y) is 1, the equation is
y<1/3x+1
BUT WAIT!
The line is solid meaning that we have to include the line as part of the values. This will make the new equation
y≤
9514 1404 393
Answer:
N = 5; only one rate is possible
Step-by-step explanation:
The values of the sets of chips are ...
12 +9N +8N^2 +4N^3
and
2 +N +N^3 +N^4
We want to find N such that the difference in value is zero:
N^4 -3N^3 -8N^2 -8N -10 = 0
A graphing calculator can show us the roots. There is only one positive real root to this equation: N = 5.
The exchange rate N is 5. Only one rate is possible.
Answer:
-63
Step-by-step explanation:
Answer:
Chessebuger
Step-by-step explanation:
see the chesse burger is real good