Solving x^2+25=0

Learn how to solve x^2+25=0, which can be done with factoring by the difference of two squares with complex numbers and also write x^2=-25 then take the square roots to both sides.
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#math #algebra #mathbasics

[TheKeller]

The second solution(subtracting 25 from 0) is immediately where my brain went, although I forgot about the +/- options and only went for 5i as a solution

[selahattinkara-oh]

quadratic equations; those who know the methods of multiplication and addition of roots can easily learn both to solve equations and to create equations.👏

[MattSuguisAsFondAsEverrr]

looks like someone didn't unlock the complex plane yet😔

[BillyOfTheDay]

I was today's years old when I found out bprp has a painting of e made out of the numbers in order

[rohangt]

x^2 + 25 goes like x^2 - (-25), then to x^2 - (5i)^2

[mirkozinauer]

x²+5²=0
{a²+b²=(a+b*i)*(a-b*i)}
x=±5*i

[qweqqqw]

The look in the thumbnail is like: WTF?

[xanderlastname]

Factoring out an exponent
Im sure thatll work fine

[diegocabrales]

Given a complex number z = a + bi, the complex square root is a function f: C --> C (here by C I mean the set of all complex numbers), denoted f(z) = √z = z^(1/2), defined as the set of all complex numbers w such that w² = z, meaning √z = z^(1/2) is multivalued. By the n-th root theorem,

z^(1/2) = |z|^(1/2){cos{[Arg(z) + 2πk]/2} + sin{[Arg(z) + 2πk]/2}i}

where

|z| = √(a² + b²)

is the modulus of z (where the square root is now the real square root, meaning it is non-negative), Arg(z) is the principal argument of z — the unique argument z makes on the complex plane which lies on the interval (-π, π], or [0, 2π), depending on the definition, but both intervals give equivalent angles — and k is either 0 or 1. In our case,

z = -25 = -25 + 0i

so

|z| = √[(-25)² + 0²] = √[(-25)²] = √(25²) = 25

and Arg(z) = π, so

√(-25) = (-25)^(1/2) = 25{cos[(π + 2πk)/2] + sin[(π + 2πk)/2]i} = 25{cos[π(1 + 2k)2] + sin[(π(1 + 2k)/2]i}

If k = 0 then 1 + 2k = 1 + 2(0) = 1 + 0 = 1 and

(-25)^(1/2) = 25[cos(π/2) + sin(π/2)i] = 25(0 + 1i) = 25(1i) = 25i

If k = 1 then 1 + 2k = 1 + 2(1) = 1 + 2 = 3 and

(-25)^(1/2) = 25[cos(3π/2) + sin(3π/2)] = 25[0 + (-1)i] = 25[(-1)i] = 25(-i) = -25i

which means (-25)^(1/2) = ±25i.

[LucaHsu-kp]

Can you do this. plz Let ABC be a triangle with AB=AC. Let O be the circumcenter of △ABC, and let D be the midpoint of BC. A circle is drawn with center O and radius OD, intersecting AB and AC at points P and Q, respectively.
Prove that ∠POQ=90 degrees.

[MmeWhy]

Day 1 of asking Bprp a question 😅:My friend send me this question : Solve using equating the coefficients:
(bx/a) − (ay/b) + a + b = 0 and bx – ay + 2ab = 0 . Find x and y😊

[Conorator]

If you are used to standard quadratic form (ax^2 + bx + c = 0), you can convert it into that and see why it won't work without i.

x^2 + 0x + 25 = 0

What adds to 0 and multiplies to a positive? Nothing that most people would think of.
You can find what would add to 0 and multiply to -25, which is 5 and -5.
We can see why 5i and -5i work here because you're multiplying the -25 by -1.
Changing the order of 5i * -5i results in 5 * -5 * i * i, or more simply, 5 * -5 * i^2.
Because we know that i^2 = -1, we know that 5 * -5 * i^2 is equal to 5 * -5 * -1, which equals positive 25.

[Friend_of_the_One-Eyed_Ladies]

I'm not especially smart, but when I can look at an equation like this and instantly know the correct answer, it makes me wonder how dumb average people really are.

[Venomous]

For any quadratic equation:
ax² + bx + c = 0

Just see if b² - 4ac is greater than 0, less than 0, or if it is exactly 0.

If greater than 0, it means that it will have 2 "real" solutions. If it is 0, then there's only 1 real solution. If it is negative, or less than 0, then there's no real solution


For example, let's take the equation in the video.

x² + 25 = 0

It can be written as
=> (1)x² + 0b + 25 = 0

So b = 0, a = 1, c = 25

So,
b² - 4ac
= (0)² - 4(1)(25)
= 0 - 100 = -100

As it is negative, it's obvious that it's not gonna have real solutions, you would have to look for complex solutions.

[FurbleBurble]

I think part of the problem is the order in which things are taught. If you know parabolic transformation rules then you can immediately see that the parabola cannot touch the x-axis, which dictates imaginary numbers are in play.

For those who don't know what I am talking about, all sections of ax^2+bx+c (where x=/=0) tell you more or less what the graph should look like. For example, "a" being positive tells you that the parabola opens up like the letter "u", whereas if "a" were negative it would open down like the letter "n", and "c" tells you how many units above or below the x-axis the vertex will be. Since "a" and "c" are both positive, the parabola will open up and never touch the x-axis, and no math is required to know this because this is always the case.

Edit: Personally, learning transformation rules was a struggle and I never could commit them all to memory. Fortunately, I never had to because my professor allowed us to use notes on tests/exams. But, simply knowing about how "a" and "c" work can potentially make life easier for people learning quadratics.

[GeoLover]

yes as (x+5)²=x²+10x+25 so (x+5)² is not equal to x²+5²...

[ugurcansayan]

This video reminded me:
Two wrongs don't make a right, --
but three rights do make a left.

[YasarYlmaz-jpyr]

Bir zamanlar masum olan ama simdi seytan dolabinin ayagi olan insanlarla abinin ve benim sonumu hazirlamak icin sende yardim et hic bos durma

[SDCORL]

Really speaking it has no solution.
OR
It has a very complex solution.

[tarapandey]

Day 3 asking for 2+2 ≠ 2x2

Source perfect definition of multiplication in zero by vigyan Darshan.