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.
27 times 3= 81 then add 9 which then equals 89
Answer:
x = 8.66
Step-by-step explanation:
Apply pythagorean theorem to find the value of x.
Thus:
x² = 14² - 11²
x² = 75
x = √75
x = 8.66025404
x = 8.66 (nearest hundredth)
Answer: x = 6.32 or -0.32
Step-by-step explanation:
2x² - 6x + 10 = 0
No we divide the expression by 2 to make the coefficient of x² equals 1
We now have
x² - 3x + 5 = 0
Now we remove 5 to the other side of the equation
x² - 3x = -5
we add to both side square of half the coefficient of x which is 3
x² - 3x + ( ⁻³/₂)² = -5 + (⁻³/₂)²
(x - ³/₂)² = -5 + ⁹/₄
Resolve into fraction
(x - ³/₂)² = ⁻¹¹/4
Take the roots of the equation
x - ³/₂ = √¹¹/₄
x - ³/₂ = √11/₂
x = ³/₂ ± 3.32/₂
= 3+ 3.32 or 3 - 3.32
= 6.32 or - 0.32
