Solve and Check Absolute Value Equation | Olympiad Math Question | Complex Algebra Problem

Can you solve and check given complicated absolute value equation for x? One side of the equation is quadratic whereas the other side is linear. absolute(x^2-49)=absolute(3x+21)

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Solve and Check Absolute Value Equation | Olympiad Math Question | Complex Algebra Problem

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[anatoliy]

Rarely in my life have I encountered abs equations. However, this is an interesting topic.Thank you for a math "raisin", Professor! Good luck and all the best to you!

[RylanceStreet]

The original equation can be expressed in two ways:
X^2 - 49 = 3x + 21 or
X^2 - 49 = - (3x + 21)
These are two quadratic equations which can be solved to give
X = -7 or 10 and
X = -7 or 4.

[davidfromstow]

I'd never heard of the term 'Absolute Value' but by doing simple quadratic equations, I got both -7 and 10; obviously I felt very pleased with myself. Pride comes before a fall! I was, therefore, intrigued to see your 3rd solution. Thank you

[sparshraj]

I love modulus because 99% of the time ...I got it right 😎😎
Great video ..loved it sir !!

[HappyFamilyOnline]

Amazing video👍
Thank you so much for sharing😊

[eddyleonelmenesespedr]

Althought this channel is in English, I can say that this video is the best

[math]

very well done, love this absolute value equation bro

[mustafizrahman]

X=-7, 4 and 10. Good question. A lot of greetings and humble respect from Bangladesh.

[ishitamondal]

Wow again a new interesting video ..
Thanks a lot sir .... 🙂🙂🙂🙂
Lots love from India ..

[mahalakshmiganapathy]

Thank you for uploading very good problems

[devondevon]

answer x = -7, 10 and 4 from:
x^2 -3x -70=0 and from
x^2 + 3x- 28
solving those two quadratic equations will give the result.

[nezamasarie]

Solved this one
Just solved this equation normally by disregarding the absolute value since
|x| when x is positive is x
|x| when x is negative is still x
i factored both eq's
so i got
(x-7)(x+7)=3(x-7)
x-7=3 therefore x=10
and we can set the eq to 0
so we get that x is -7
to get x=4
we can use the properties of absolute values
and use -3 instead of 3 on the left hand side
giving us that 4 is a valid solution

[barryday]

Just square both sides to remove the modulus signs then factorise.

[АлександрИванов-фэя]

Возвести обе части в квадрат. Избавиться от модулей. Разложить как разность квадратов и решить два квадратных уравнения. Задача решается в уме

[marklevin]

Original equation implies that either x^2-49=3x+21 or
X^2-49= -(3x+21)
Those are quadratic equations can be solved with quadratic formula
First has roots -7 and 10
Second has roots -7 and 4.
Because -7 is a common root of both equations we only get 3 solutions ... -7, 4, 10

[luizmiguel]

Norm of a Vector under Real Vector Space (R¹): "||û||:= ||û|| = |û| = √(x-x₀)² ; û ∈ û¹ or R¹ (Real Vector Space)". It can be useful in Limit's Analysis.

[Gavin_Gladiator]

My answer before he started explaining. -7

I was a third correct

[A--VoAnhTuan]

If x belongs to the set of real numbers then the answer on the video is correct, otherwise the correct answer will be x=-7 and |z-7|=3, where z the set of complex numbers belongs to the circle with center (7, 0) and the radius is 3.

[alster]

Alternative way of getting -7 and 10 is solving the quadratic equation itself

x²-3x-70=0

(x-10)(x+7)


x= 10, -7 then then 4 is also easy to find.