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:
The least common multiple (LCM) of two or more non-zero whole numbers is the smallest whole number that is divisible by each of those numbers. In other words, the LCM is the smallest number that all of the numbers divide into evenly.
Step-by-step explanation:
The LCM of 13 AND 17 is 221
The LCM of 40 AND 60 is 120
lcm (40; 60) = 120 = 23 × 3 × 5
Answer:
The length is 14 units
Step-by-step explanation:
Take the value at point F and subtract the value at point E
F = 19
E = 5
19 -5 =14
The length is 14 units
Answer:
18 1/2 years old
Step-by-step explanation:
x-3= 15 1/2
x= 15 1/2 + 3
x= 18 1/2
Hunter is 18 1/2 years old.
2/5+3/8= 31/40 therefore he still has 9/40 to go.
to get this you multiply the two denominators together
5x8=40 which is your denominator
then you cross multiply the two numerators
2x8=16
3x5=15
to get the final
15/40+16/40=31/40
and you are left with 9/40
