The correct answer is C
Good luck on the rest :D
Answer:
![$\[x^2 + 22x + 121\]$](https://tex.z-dn.net/?f=%24%5C%5Bx%5E2%20%2B%2022x%20%2B%20121%5C%5D%24)
Step-by-step explanation:
Given
![$\[x^2 + 22x + \underline{~~~~}.\]$](https://tex.z-dn.net/?f=%24%5C%5Bx%5E2%20%2B%2022x%20%2B%20%5Cunderline%7B~~~~%7D.%5C%5D%24)
Required
Fill in the gap
Represent the blank with k
![$\[x^2 + 22x + k\]$](https://tex.z-dn.net/?f=%24%5C%5Bx%5E2%20%2B%2022x%20%2B%20k%5C%5D%24)
Solving for k...
To do this, we start by getting the coefficient of x
Coefficient of x = 22
<em />
Divide the coefficient by 2
Take the square of this result, to give k
Substitute 121 for k
The expression can be factorized as follows;
<em>Hence, the quadratic expression is </em><em></em>
Here, we can split this parallelogram into two triangles and one rectangle. The triangles are equal to each other so we can just find the area of the rectangle and the are of <span>one triangle.
</span>Triangle area formula:
, where a = area, b=base, and h=height.
Plug in 5 for b and 11 for h:
<span>Multiply by 2!
</span>
Rectangle area formula: a=bh, where a = area, b = base, and h = height.
Plug in 4 for b and 11 for h.
a = 4 x 11
a = 44
44+55 = 99
Area is 99 cm<span>².</span>
It bears repeating: you should have applied the well-known following formulas: "Hello", "Please" and "Thanks".
Anyway...
= = = = = = = = = = = = = = = = = =
The way you stated the problem, there is an infinity of possibilities for the other solution.
► For instance, the quadratic equation:
x² – (6 + 4i)x + (9 + 12i) = 0
has for discriminant:
Δ = (6 + 4i)² – 4(9 + 12i) = 36 – 16 + 48i – 36 – 48i = -16
which is indeed negative.
Its solutions will then be:
x₁ = [(6 + 4i) + 4i]/2 = 3 + 4i
x₂ = [(6 + 4i) – 4i]/2 = 3
And the other solution here is 3.
► If you are not convinced, the quadratic equation:
x² – (6 + 5i)x + (5 + 15i) = 0
has for discriminant:
Δ = (6 + 5i)² – 4(5 + 15i) = 36 – 25 + 60i – 20 – 60i = -9
which is indeed negative.
Its solutions will then be:
x₁ = [(6 + 5i) + 3i]/2 = 3 + 4i
x₂ = [(6 + 5i) – 3i]/2 = 3 + i
And the other solution here is 3+i.
► In fact, every quadratic equation of the form:
x² – [6 + (4 + α)i]x + (3 + 4i)(3 + αi) = 0
where α is any real, has for discriminant:
Δ = [6 + (4 + α)i]² – 4(3 + 4i)(3 + αi)
= 36 – (4 + α)² + 12(4 + α)i – 36 + 16α – 12(4 + α)i
= 16α – (4 + α)²
= 16α – 16 – 8α – α²
= -16 + 8α – α²
= -(α – 4)²
WILL be negative.
Their solutions will then be:
x₁ = [ [6 + (4 + α)i] – (α – 4)i ]/2 = 3 + 4i
x₂ = [ [6 + (4 + α)i] + (α – 4)i ]/2 = 3 + αi
And the other solution will then be is 3+αi.
Since α can take any real value, you'll obtain an infinity of solutions of the form 3+αi.
► So conclusively:
If the discriminant of a quadratic is negative AND one of the solutions is 3+4i, the only thing we can say about the other solution is that its real part must be 3.
Regards, to lizard squad >:0
