The answer is <span>A –x^2 + 3x + 15
</span>
3x² + 6 – 2x + 5x – 4x²<span> + 9
Rearrange it:
</span>3x² + 6 – 2x + 5x – 4x² + 9
= 3x² - 4x² - 2x + 5x + 6 + 9
= -x² + 3x + 15
Answer:
just one.. so A
Step-by-step explanation:
Well if the number is 100 dollars and it increases by four hundred percent it now costs 400 dollars. But if it was 100 dollars after the increase the original price was 25 dollars.
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.
