Can these equations have imaginary solutions?

Question

(y + 12)^2 + 400 = 0
(x - 5)^2 + 100 = 0
(y + 12)^2 - 400 = 0

I do not think they can, becuase they are missing the middle number with the X

EX: 2x^2 + 4x + 16 = 0
Aren't all of the above equations missing the number where 4x is on this example?

If I am wrong, can you please explain? Thank you.

P.S. You do not have to solve any of these.

Nevermind. I figured it out.

But since someone has answered, I will keep this up so someone can get 10 points.

Answer 1

So, these can have imaginary solutions. Just because they are missing the X doesn't mean much, it just means you factor them DIFFERENTLY.

For example: (2x+1)(2x-1) = 4x^2-1 It's also missing the middle X term, (and granted it's not imaginary) but it is still factorizable and has two solutions, +1/2 and -1/2


I'll post the answers to the equations a few lines down if you want to read them to check your work:




(y+12)^2 = -400=> y+12 = 20i or -20i
=> y= 20i-12 or -20i-12

(x-5)^2 = -100 => x-5 = 10i or -10i
=> x=5+10i or 5-10i

(y+12)^2 = 400
y+12 = 20 or -20
y = 8 or -32

In the last case, the solution is not imaginary.

Answer 2

The first two have imaginary solutions, the third one has a real solution.
An equation does not have to have a linear term in order to have a solution. [("the middle number with the x") = linear term]

I will give you a hint. Do you know how to factor a^2 - b^2 = 0?
(a+b)(a-b) = 0

Also remember, +400 = -400i^2, since i^2 = -1

(y+12+20i)(y+12-20i)=0
(x-5+10i)(x-5-10i)=0
(y+12+20)(y+12-20)=0

Answer 3

First, they all have solutions if you allow for imaginary roots.

The first one has only imaginary roots because you can rewrite it as: (y + 12)^2 = -400 or a square = a negative. Only imaginary roots. You would get: (y + 12) = +/- 20i, y = -12 +/- 20i

Same thing for the second one: (x - 5)^2 = -100 so
x-5 = +/- 10i, x = 5 +/- 10i

The last one has real roots: (y+12)^2 = 400
y + 12 = +/- 20, so y = -12 +/-20 or, 8 and -32.

Answer 4

The first two have imaginary solutions the third doesnt.

Your forgetting to expand if you want it it in the form Ax^2 + Bx + C
You would need to expand and collect like terms to be in that form.

But the easy way to see it is that its in this form

z^2 + b = 0 (edit - also note that for all z; z^2 > 0)

if b > 0 then z = isqrt(b) is the imaginary solution

if b <= 0 then z = -sqrt(b) and z = sqrt(b) are the solutions.

Answer 5

(y + 12)^2 + 400 = 0
(y + 12)^2 = -400
y + 12 = ±20i
y = -12 ± 20i

(x - 5)^2 + 100 = 0
(x - 5)^2 = -100
x - 5 = ±10i
x = 5 ± 10i

(y + 12)^2 - 400 = 0
(y + 12)^2 = 400
y + 12 = ±20
y = 8 or -32

---------------------------------

2x^2 + 4x + 16 = 0
2(x^2 + 2x + 8) = 0

x = (-b ± sqrt(b^2 - 4ac))/(2a)

x = (-2 ± sqrt(2^2 - 4(1)(8)))/(2(1))
x = (-2 ± sqrt(4 - 24))/2
x = (-2 ± sqrt(-20))/2
x = (-2 ± sqrt(-4 * 5))/2
x = (-2 ± 2isqrt(5))/2
x = -1 ± isqrt(5)

Answer 6

sure they have a middle term. here's how:
(y + 12)^2 + 400 = 0
y^2 + 24y + 400 = 0
y^2 + 24y + ___ = - 400 + ___
y^2 + 24y + 144 = -400 + 144
(y + 12)^2 = -256
y + 12 = plus or minus sqrt -256
y + 12 = plus or minus 16
y = -12 plus or minus 16
therefore, y = -28 or y = 4
for this one there is no imaginary part
this is how I'd do the problem, the rest are yours to do

Answer 7

a) attempt to take the sq. root as you frequently might. submit to in techniques that it has a great and a detrimental sq. root. w = ±sqrt(-225) component -225 into -a million*225 consequently w = ±sqrt(225)*sqrt(-a million) the sqrt(225) = 15. The sqrt(-a million) = i, via definition of i in case you call it like a mathematician. effect: w = 15*i or w = -15*i --------------------------------- b) 3*v^2 + 4*v - a million = 0 Use the quadratic formulation: v = (-4 ± sqrt(4^2 - 4*(-a million)*3))/(2*3) simplify: v = -2/3 ± sqrt(7)/3 consequently: v = sqrt(7)/3 - 2/3 or v = -2/3 - sqrt(7)/3

Answer 8

When you FOIL out the parentheses (y+12)(y+12)= y^2 + 24y + 144, you will have that middle term you need.