Nice Radical Math problem | Find value of ‘a’

[@georgewu5885]

The original equation is equivalent to (1+i)√a=12; multiplying both sides by (1-i) yields 2√a=12(1-i), or √a=6(1-i) thus a=36(1-i)^2=-72i;

[@Frank-kx4hc]

Sqrt(-a)=s*is*qrt(a), s=+1 or -1 Thus sqrt(a)=12/(1+i*s) =12(1-is)/(1+s^2)=6(1-is) a=36(1-s^2-2is)=72i or -72i

[@martinzavalaleon8856]

@georgewu5885, está claro que no son equivalentes porque la ecuación original tiene soluciones a pares y tú obtuviste solo una solución. También dirías que la ecuación original es equivalente a (1+i)sqrt(-a) = 12 y obtendrias otra solución. ¿Dirías que (1+i)sqrt(a) = 12 y (1+i)sqrt(-a) = 12 son equivalentes?

[@sergeygaevoy6422]

sqrt(a) + sqrt(-a) = 12 (sqrt(a) + sqrt(-a))^2 = 12^2 a + 2a * sqrt(-1) + (- a) = 144 a= 72 / sqrt(-1) i^2 = (-i)^2 = -1 => a = +- 72i

[@jahangirtamboli477]

What is the solution significance? Mere mathematics.

[@Frank-kx4hc]

Sqrt(-a)=s*i*qrt(a), s=+1 or -1 Thus sqrt(a)=12/(1+i*s) =12(1-is)/(1+s^2)=6(1-is) a=36(1-s^2-2is)=72i or -72i

[@allanflippin2453]

That's an interesting problem! Can you tell me how to check the results? I've tried, but I'm not getting "12" after the math. Thanks. It's my fault because I don't know how to deal with sqr(-1) inside a sqr.

[@joeviolet4185]

You are right. sqrt(72·i) + sqrt(-72·i) = [sqrt(72)/sqrt(2) + i·sqrt(72)/sqrt(2)] + [-sqrt(72)/sqrt(2) + i·sqrt(72)/sqrt(2)] = 2·i·sqrt(36) = 12·i, but not Real 12. Thus, I need a solution, where the Imaginary parts of the Complex roots cancel out and not the Real parts. The error in calculation is that i^2 = -1, but i^6 = -1, too. The double squaring in finding the solutions makes it necessary to multiply the solutions with i². When I check the initial equation with = 72·i³ and a = -72·i³ I get the answer 12, because then taking the square roots leads to sqrt(a) = sqrt(72·i³) = i·sqrt(72·i), which not only exchanges Real and Imaginary Parts of the complex number, but also multiplies the former Imaginary part with -1, i.e., changes the sign of the resulting Real part. In this case, the Imaginary parts of the roots cancel out and the sum of the Real parts is a Real number.

[@allanflippin2453]

@@joeviolet4185 Thanks! I know I couldn't figure it out anyway.

[@rbtmdl]

Where'd you get x and q? There wasn't any x and q in the original formula then all of a sudden we have an x and a q.

[@anatolykhmelnitsky2841]

√a + √(-a) = 12; √a +√(-1*a) = 12; √a +i√a = 12; √a(1 + i) = 12; √a = 12/(1 + i); (√a)² = (12/(1 + i))²; a = 144/(1 + i)²; a = 144/(1 + 2i -1); a = 144/2i; a = 72/i; a = 72i/i²; a = -72i; ±a = ∓72i; a₁ = -72i; a₂ = 72i;

[@Frank-kx4hc]

@@anatolykhmelnitsky2841 a=144/(2i) = -72i In the begining sqrt(-a)=+or-isqrt(a) Thus a=+or-i72 .

[@ForestHills101]

a=-72i is the only solution (real radical). a=+72i is not a solution (imaginary radical).

[@Frank-kx4hc]

a=+ or- 72i because sqrt(-a)=+or-isqrt(a)

[@rafaelyorro4466]

Lo hiceeee mas corto -72

[@Frank-kx4hc]

Very bad ! Return to school

[@ajeethpandey]

Irritating way of explanation !

[@송세흠]

This problem is not right. We can not have negative number in a route. One of a or -a is negative number.

[@paulokas69]

In complex numbers, we can have sqroot of negative numbers

[@anoopkumar-dt7wp]

Understood that You are just a kid. When you reach higher classes, you will learn the concept of complex numbers...😊