A beautiful math problem for advanced students | Equation Solving

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Equation: a^3 + a^2 = 36

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[salehawadh]

Soln

a³+a²=36
a³+a²=27+9
a³+a²=3³+3²
a³=3³, a²=3²
a=3 answer

[rogerphelps]

Writting 36 as 27+9 presupposes you know the answer.

[gibbogle]

A trivially simple problem for beginners. a=3 by inspection

[reneeshcr]

a³+a² =36
a²(a+1)=36
a²(a+1)=9×4
a²>(a+1)
By way of comparison,
a²=9 and a+1=4
a=√9 and a=4–1
a=3 and a=3
Finally, a=3.
It's just a different perspective.

[HoSza]

a=1? No... a=2? No... a=3? Yes!!! Let's divide a³+a²-36 ÷ a-3 = a²+4a+12. So the other two solutions: 2a=-4±√-32, two more complex roots a=-2+i√8 and -2-i√8.

[chrisclub]

To really solve this just write a^2(a + 1) and notice that if a were to be an integer, it’s square must divide 36, and the only squares that divide 36 are 4, 9, and 36, so a must be 2, 3, or 6, from there, check

[Kndiani]

This method of solving the equation begins by guessing the correct answer! Why would anyone even continue once you got 27 & 9? That's your answer right there.

[MgtowRubicon]

a^3 + a^2 = (a^2)×(a + 1) = (3^2)×(3 + 1) = 9×4 = 36

[freddyalvaradamaranon]

Muy interesante video sobre ecuaciones con exponentes cuadrados y cubos, muchas gracias por compartir.😊❤😊.

[narottamdeep]

3 to the power 3 + 3 to the power 2 is 36. Hence value of a=3

[asf]

Just replace 'a' by '1/b' to cast the equation in standard form then use Cardano (Tartaglia) equation to find one root and then divide by (b-root) to find the other two!

[Ron_DeForest]

Awesome explanation. If you went with just the obvious answer you wouldn’t have found the imaginary numbers. What I want to know is why hasn’t anyone commented about the most obvious part of the video. Your crazy good hand writing. I’d love to be that legible.

[rosbelisoropeza]

Ok. La idea es pensar. Usar los productos notables. Como tarea debe ser direccionada y contestar con explicaciones en cada proceso . Así un estudiante mejora su capacidad cognitiva.

[dneary]

If you are given that the solution is in integers, then a^2(a+1) = 36 means a is in {1, 2, 3, 6} - and process of elimination gives an answer. If you do not know that a is an integer, then you can try to identify where there are roots to a^3 + a^2 - 36 = 0 and you quickly get (a - 3)(a^2 + 4a + 12) = 0, or a = 3, -2 +/- 4✓(-2)

[hemrajparihar]

a = 3
3 raised to 3- 27
27 + 3 raised to 2
27 +9 - 36

[EnginAtik]

Re-write the problem as a^3=(6-a)(6+a) right hand side should be factorable by a therefore 6-a = 3 => a = 3.

[rizanomurni]

If determinant D < 0, it mean the equation nothing the root.

[prollysine]

by faktoring, a^3+a^2 |+/-| n*a-36=0, / 3*12=36, trial n=3 /, a^3-3a^2+4a^2-12a+12a-36=0, (a-3)(a^2+4a+12)=0, a-3=0, a=3,
/// for complex, a^2+4a+12=0, ///, test, 3^2+3^2=27+9, 27+9=36, OK,
a^2+4a+12=0, a=(-4 |+/-| sqrt(16--48)/2, a=(-4 |+/-| sqrt(-32))/2, a=(-4 |+/-| sqrt(32)i)/2, a=(-4 |+/-| 2*sqrt(8)i)/2,
a= -2+sqrt(8)i, -2-sqrt(8)i, solu., a= 3, -2+sqrt(8)i, -2-sqrt(8)i,

[ressouguerrier]

a³+a²=36; solution;a³+a²=36; a³+a²-36=0; a³+a²-27-9=0;a³-27+a²-9=0; a³-3³+a²-3²=0; (a-3)(a²+3²+3a)+(a-3)(a+3)=0; (a-3)[a²+9+3a+a+3]=0; a-3=0; a=3;a²+12+4a=0; a²+a+12=0; ∆=4²-4×1×12; ∆=-32; a=3; S={3}.

[bandarusatyanandachary]

It can be solved by using reminder theorem in a simplified way.