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:
2/6
Step-by-step explanation:
because 4/6 is eaten so that means 2/6 is left to eat
Answer:
Potassium Iodide
Step-by-step explanation:
KI (potassium iodide) is a salt of stable (not radioactive) iodine that can help block radioactive iodine from being absorbed by the thyroid gland, thus protecting this gland from radiation injury. The thyroid gland is the part of the body that is most sensitive to radioactive iodine.
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Answer:I just need points
Step-by-step explanation:
Hey
35-15 i think dont hate me if im wrong
