## Olympiad Mathematics | Equation Solving | You should use this trick

Thanks! Share it with your friends! How to solve this equation? By using this trick, you can deal it quickly! Bert Bril says:

Once you find '3', a fun way to find the resulting quadratic equation to solve for the other two roots is … long division. Just set up a long division for /a^3+a^2-36a-3. First round: a^2, gives a^3-3a^2, subtract it to leave 4a^2-36. That goes 4a times, leaving 12a-36, etc until leaving rest 0. The new equation is a^2+4a+12, hence a^3+a^2-36 can be written as (a-3)(a^2+4a+12) . Gautam Kumar says:

We can solve a=3 no need to assume aditya gamer says:

Directly my mind say x=3 Alex Op says:

Oooh, so you must first guess roots and then solve an equation. I see OK Play says:

I guess you can use the cube root equation, but who even remembers it. Plus you might not get such a simple looking answer. Pinaki Roy says:

Lady mathematical problems are not solved by guesswork. It takes you 6 minutes. Come to India. Girolamo Capita says:

4*9=36 3^2=9. 3+1=4. (A^2)*(A+1)=36 堀勇作 says:

a^3+a^2=36 =27+9 堀勇作 says:

a^3+a^2=36=27+9
ゆえにa=3 242 MATH says:

very well done, thanks for sharing Subbarao B says:

Write perpect spures and క్యూబస్
14,9,16,25ఎత్చ్
1,8,27
సొల్యూషన్ 27+9=36 Laszlo Liptak says:

In my opinion, once you have a root, doing long division is simpler than going through all that factoring. David Brisbane says:

Let's see if we can solve this equation by finding a real root of the equation using the way the cubic root of such equations was first discovered.

a³ + a² – 36 = 0 is a cubic, and so it has at least one real root.

A bit more investigation uncovers that it has only one real root, and that this real root is positive.

First we will reduce the equation to its depressed form, which will remove the square term from the equation.

Let a = x – 1/3.

a = x – 1/3 and a³ + a² – 36 = 0
⇒ (x – 1/3)³ + (x – 1/3)² – 36 = 0
⇒ x³ – x/3 – 970/27 = 0.

This is the depressed cubic.

This depressed cubic also has at least one real root. In fact it only has one real root.

Let this real root be x and let x = u + v, where u and v are real numbers. Introducing u and v like this is the insightful part of the approach, which eluded mathematicians for the good part of two thousand years.

Now x = u + v and x³ – x/3 – 970/27
⇒ (u + v)³ – (u + v)/3 – 970/27 = 0
⇒ (u³ + 3u²v + 3uv² + v³) – (u + v)/3 – 970/27 = 0
⇒ (u³ + v³ – 970/27) + (u + v)(3uv – 1/3) = 0

So, if we can find u & v such that
(u³ + v³ = 970/27) and (3uv – 1/3 = 0), then we can find x, and hence a.

Solving u³ + v³ = 970/27 and 3uv = 1/3
⇒ u³ + 1/(9³u³) = 970/27
⇒ 729u⁶ – 26,190u³ + 1 = 0

This is a quadratic equation in u³.

⇒ u³ = [26,190 ± √(685,913,184)]/1,458
⇒ u = ³√[(26,190 ± √(685,913,184))/1,458]

Let's choose the -ve square root (if we had chosen the +ve square root, we'll get the same result for x).

u = ³√[(26,190 – √(685,913,184))/1,458] & v = 1/9u
⇒ v = (1/9) * ³√[1,458/(26,190 – √(685,913,184))]

Hence x = u + v =
³√[(26,190 – √(685,913,184))/1,458] +
(1/9) * ³√[1,458/(26,190 – √(685,913,184))]

We can compute this with a calculator and find the result is x = 3.333333 (6 decimal places).

However, we can do horrendous algebra to deduce that
x = u + v =
³√[(26,190 – √(685,913,184))/1,458] +
(1/9) * ³√[1,458/(26,190 – √(685,913,184)) = 10/3

Now as a = x – 1/3 and x = 10/3
⇒ a = 3.

We can avoid the horrendous algebra by reversing the logical steps to show that the horrendous equation satisfies x³ – x/3 – 970/27 = 0 and a³ + a² – 36 = 0, and by inspection a = 3 is the real root, so x = 10/3. However, using the reverse logical steps approach means that determining the exact root by the method demonstrated above is moot.

However, if the original equation were a³ + a² – 35 = 0, then the above methods would find the exact real root albeit in the form of cube and square root combinations, whereas a factorisation approach would not find the real root.

There is extra algebra about complex roots that will allow the other two roots to be found by a simple complex number manipulation of the real root. hafedh saif says:

9*4
9*(3+1)
3^3*(3+1)
a=3 Aisha sario says:

I Solve your any type of math question on fiverr albert figueres giral says:

The value for a is 3. Also we can factor out a^2 because it's a common term in both numbers. Therefore we'd have this: a^2·(a+1)=36
Substitute the value of 3 in this ecuation and we have 9·4, which is equal to 36 이호석 says:

Thanks ma'am Vixlife says:

You make a blocking hell lotta asumptions, ya know right? Kartik Kale says:

By observation, a=3 Needles says:

It depends on how we're asked to solve the problem. One way is to realize that the greatest common factor on the left hand side of the problem is a^2. a^2(a+1)=36 -> a = 36/a^2 – 1. For 1, 1 = 35. This is false. For 2, 2 = 8. This is also false. Lastly, 3 = 3. This is true. Knowing this we can then realize that while factoring a^3 + a^3 = 36 we can start with (a-3). Divide a^3+a^2-36 by (a-3). This leaves us with a^2 + 4a + 12. This is easier to factor. Using quadratic formula you can now find the other solutions. x=(-b+/-(b^2-4ac)^.5)/2a. Hence x=(-4+/-(16-4(1)(12))^.5)/(2(1)). This gives us -2+/-2i(2^.5) for our other 2 possible solutions. Let's prove it. Plus these solutions back into the original equation. When dealing with -2+2i(2^.5) a^2 = -8i(2^.5)-4. a^3 = 16i(2^.5)+8+16(2)-8i(2^.5). Add a^2 and a^3 to get 36. Do the same with the other solution. I'll leave this for you to prove. Alok Verma says:

Imaginary roots will be -2+/- 3i Roliver says:

a³ + a² = 36

a².a + a² = 9.4

a²(a+1) = 9.4

a² = 9
a = ±3

a+1 = 4
a = 3

→ a = 3