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.
Step-by-step explanation:
a = 5
b = -3
25 - -3/ -3 - 25
= 28 / -28
= -1
15/2 if you are binging it to simplest form
Answer:
Step-by-step explanation:
r6tygtf77777777777777777777777777777777777777777777777777777777777777777
Answer:
so you would rewrite them as (2x-5)(3x-4)
then multiply 2x×3x=6x^2 then -5×-4=20
so the expression is C. 6x^2+20
