M² - M³ = 150 M²(1 - M) = 25(6) ---------1️⃣ 1 - M = 6 M = -5✅ Sub the value of M in equation 1️⃣, (-5)²(1-(-5)) = 25(6) 25(6) = 25(6) 150 = 150 Hence,proved💯
This is easy to do in your head. Once you realize that 5² = 25 and 5³ = 125 and 25 + 125 = 150, you wonder how you can put them together. Since (-5)³ = -125, the answer HAS to be -5!
Since we have a cubic function, there must be 3 solutions. Hence only deriving M = - 5 is not a complete answer. Other two solutions have to be worked out.
It has to be a negative. A neg^2 is positive and a neg^3 is negative. Makes sense. Try -2. (-2)^2 - (-2^3)= 4 + * = 12 Try -3. 3^2 + 3^3 = 36 Try -5. 5^2 + 5^3 = 25 + 125 = 150
Let M^2=a. Then the equation becomes a-a√a=150. Now a-150=a√a. Squaring both sides we get (a-150)^2=a^3. By bringing RHS to left side. You (a-150)^2=a^3. You have a^3-a^2+22500=0. It is a cubic equation . You can get the value of a. Then root of that gives you the value of M.
First of all her, I appreciate her method of doing. As it is a bit advanced problem, a training must be given in the following way. Clearly M not equal to 0 and +1 & -1 (verify by substitution. Also M cannot be between -1 and +1 as RHS is 150. So M must be either > 1 or < -1. Once again if M > 1 cannot satisfy the problem by substituting any value > 1. Hence M < -1. NOTE THAT ALL THE ABOVE EXPLANATION CAN BE DONE EASILY IN THE BACK OF ONE’S MIND. Not a big deal. A training towards mathematics must be like that. Now our answer is M < -1. Now guess work starts (also called inspection). 150 = 6x5^2 or 125 +25 =5^3 +5^2 --> M = -5 Please accept her method also. 👍
You don't understand why DELTA is calculated. In your solution, delta is less than zero, which means that the equation has no solutions, and you want to prove otherwise by force. I don't think you understand the result of delta. Delta is calculated to see if the equation has solutions. When delta is less than zero, it means that the equation has no solutions, because the graph of the function does not intersect the X axis. After all, you equated the equation to 0, so Y has the value 0, which means that only those x are solutions that touch the x axis. And since delta is negative, there is no x that would touch the x axis.
Why complicate like this? It is very simple. When M+5 multiplied by something is 0, take that something to the other side, the value becomes 0 (0 divided by anything is 0). Hence M+5 is 0 and hence M is -5.
This is a third grade equation not an exponential one. However it wasn't specified the field number applied so the second grade equation couldn't be calculated in the real field but only in complex one.
M^2 -M^3 = 150 , M= 3+√21 i, (3+√21i)^2 - (3+√21i)^3 = 150 if we simplify this equation we will get (a+b)^2=a^2+b^2+2ab (a+b)^3= a^3+b^3 +3ab(a+b) 9-21+6√21i -27 +21√21i -27√21i+189 = 150 , 198-48+27√21i-27√21i =150 150=150 , you can also check by yourself, then you will know is this right or not.
Apply M = 5 in the given equation, you won't get tallied on both sides. M²-M³ = 5² - 5³ = minus 120 which is not equal to 150 as given in question Hence 5 is wrong and minus 5 is correct answer.
Se ai miei tempi ci avessi così tanto e così pesantemente a risolvere questo stupido problemino il professore mi avrebbe come minimo dato solo la sufficienza
M^2 -M^3 = 150 , M= 3+√21 i, (3+√21i)^2 - (3+√21i)^3 = 150 if we simplify this equation we will get (a+b)^2=a^2+b^2+2ab (a+b)^3= a^3+b^3 +3ab(a+b) 9-21+6√21i -27 +21√21i -27√21i+189 = 150 , 198-48+27√21i-27√21i =150 150=150 , you can also check by yourself, then you will know is this right or not.
m^2-m^3=150 m*m*(1-m)=2*3*5*5 m*m*(m-1)=-5*5*6 m*m*(m-1)=(-5)*(-5)*(-6) m*m*(m-1)=(-5)*(-5)*(-5-1) m=-5 is a solution. m^3-m^2+150=0 (m+5)*(m^2-6m+30)=0 m=-5 or m^2+6m+30=0 m=-5 or m=-3+-sqrt(3^2-30) since 9-30 is negativ, there is nooter real solution.
We can solve this problem by using Newton- Raphson method, follow these steps: 1. Rewrite the Equation First, we rewrite the equation as: f(M) = M^2 - M^3 - 149 = 0 2. Compute the Derivative Next, compute the derivative of f'(M) = 2M - 3M^2 3.The Newton-Raphson iteration formula is: M_{n+1} = M_n - {f(M_n)}/{f'(M_n)} Substitute ( f(M) ) and ( f'(M) ) into the formula: M_{n+1} = M_n - {M_n^2 - M_n^3 - 149}/{2M_n - 3M_n^2} 4. Initial Guess Choose an initial guess for ( M_0 ). You can start with a reasonable value like ( M_0 = 5 ), but the closer the initial guess is to the actual root, the faster the convergence Let's perform the calculation step by step. I'll do the iteration using an initial guess of ( M_0 = 5 ) The Newton-Raphson method converged to a solution ( M approx -4.9882 ) after 12 iterations. This value of ( M ) approximately satisfies the equation ( M^2 - M^3 = 149 ). Here are the 12 iterations of the Newton-Raphson method M_n M_{n+1} 5.0 1.1692 1.1692. -83.4847 -83.4847 -55.5533 -55.5533 -36.9416 -36.9416 -24.5543 -24.5543 -16.3416 -16.3416 -10.9663 -10.9669 -7.59547 -7.59547 -5.75294 -5.75294 -5.08055 -5.08055 -4.98978 -4.98978 -4.988210 After 12 iterations, the method converges to a solution of approximately ( M approx -4.9882 ). M~ -4.9882 Now let's verify M^2 - M^3 =149 (-4.9882)^2 -(-4.9882)^3 = 149 24.882 - ( -124.117) = 149 24.882+124.117= 149 148.999 = 149 149 =149 So m = -4.9882 .
To solve for M, we'll use algebra. Given equation: M² - M³ = 150 Rearrange: M³ - M² + 150 = 0 Factor out M²: M²(M - 1) + 150 = 0 Subtract 150 from both sides: M²(M - 1) = -150 Divide both sides by (M - 1): M² = -150 / (M - 1) Now, solve for M: M² = 150 / (1 - M) M = ±√[150 / (1 - M)] Since M must be real, let's try integer values: M = 5 satisfies the equation. Verify: M² - M³ = 5² - 5³ = 25 - 125 = -100 (not 150, retry) M = 6 satisfies the equation. Verify: M² - M³ = 6² - 6³ = 36 - 216 = -180 (not 150, retry) M = 5.5 or decimal values may satisfy the equation, but let's check integers. M = 10 satisfies the equation. Verify: M² - M³ = 10² - 10³ = 100 - 1000 = -900 (not 150, retry) After retrying, we find: M = 6.7 or M = -5.5 (approximate solutions) However, exact integer solutions are challenging. Consider decimal approximations. Numerical methods or graphing tools can provide precise solutions.
M² - M³ = 150
M²(1 - M) = 25(6) ---------1️⃣
1 - M = 6
M = -5✅
Sub the value of M in equation 1️⃣,
(-5)²(1-(-5)) = 25(6)
25(6) = 25(6)
150 = 150
Hence,proved💯
I agree but -5 is one of the root.other 2 roots lies in imaginary set
What is this logic if 1-m=6 ok and what about mpower2 is it equal to 25?
@@rushyareddymaligireddy5257 Value of M = -5
So (-5)² = 25
1 - M = 6
1 - (-5) = 6
6 = 6
This is easy to do in your head. Once you realize that 5² = 25 and 5³ = 125 and 25 + 125 = 150, you wonder how you can put them together. Since (-5)³ = -125, the answer HAS to be -5!
This is simpler!
444e4444
This is so much simpler than the 🐂S imaginary formula used in the video
Lemme give a respect for best shortcut
Amazing 😂
This is wrong.
Since we have a cubic function, there must be 3 solutions. Hence only deriving M = - 5 is not a complete answer. Other two solutions have to be worked out.
brother roots can be same. for example (x-2)^2 roots are same but it's a quadratic equation. Don't even know the basics???🐖
Other two solutions are imaginary. I've done it
Another Solution
M³-M²+150=0
(M+5)(M²-6M+30)=0
※By factor theorem
M+5=0 ⇔ M=-5
M²-6M+30=0
↓
※Completing the square
(M-3)²-9+30=0
(M-3)²=-21
M-3=±√21 i
M=3±√21 i
Answer
M=-5 , 3±√21 i
If you work out the sum in such a long process in a competitive exam you imagine where you stand.
It has to be a negative. A neg^2 is positive and a neg^3 is negative. Makes sense.
Try -2. (-2)^2 - (-2^3)= 4 + * = 12
Try -3. 3^2 + 3^3 = 36
Try -5. 5^2 + 5^3 = 25 + 125 = 150
m^3 - m ^2 + 150 = 0
m^3 + 125 -m^2 + 25 = 0
( m + 5) ( m^2 - 5 m + 25 - m + 5) = 0
(m + 5) ( m^2 - 6 m + 30) = 0
only feasible solution in real domain is m = - 5
Times like this make me glad I know the rational zeros theorem and synthetic division.
Simple way:
(m^2)(1-m)=(25)(6), m^2=25, m=+/- 5, 1-m=6, -m=6-1, m=-5
Let M^2=a. Then the equation becomes a-a√a=150. Now
a-150=a√a. Squaring both sides we get (a-150)^2=a^3. By bringing RHS to left side. You (a-150)^2=a^3. You have a^3-a^2+22500=0. It is a cubic equation . You can get the value of a. Then root of that gives you the value of M.
Solving a cubic equation is more tough
First of all her, I appreciate her method of doing. As it is a bit advanced problem, a training must be given in the following way. Clearly M not equal to 0 and +1 & -1 (verify by substitution. Also M cannot be between -1 and +1 as RHS is 150. So M must be either > 1 or < -1. Once again if M > 1 cannot satisfy the problem by substituting any value > 1. Hence M < -1. NOTE THAT ALL THE ABOVE EXPLANATION CAN BE DONE EASILY IN THE BACK OF ONE’S MIND. Not a big deal. A training towards mathematics must be like that.
Now our answer is M < -1. Now guess work starts (also called inspection). 150 = 6x5^2 or 125 +25 =5^3 +5^2 --> M = -5
Please accept her method also. 👍
You don't understand why DELTA is calculated. In your solution, delta is less than zero, which means that the equation has no solutions, and you want to prove otherwise by force. I don't think you understand the result of delta. Delta is calculated to see if the equation has solutions. When delta is less than zero, it means that the equation has no solutions, because the graph of the function does not intersect the X axis. After all, you equated the equation to 0, so Y has the value 0, which means that only those x are solutions that touch the x axis. And since delta is negative, there is no x that would touch the x axis.
By inspection method and solving we get m=-5
I saw the solution -5 by inspection.
Correct, answer is simple i.e. -5.
Sol. (-5) square - (-5) cube = 25 + 125 = 150. Solved.
Why complicate like this? It is very simple. When M+5 multiplied by something is 0, take that something to the other side, the value becomes 0 (0 divided by anything is 0). Hence M+5 is 0 and hence M is -5.
Last part of this solution make with complex, it can do by the middle term method
This is a third grade equation not an exponential one. However it wasn't specified the field number applied so the second grade equation couldn't be calculated in the real field but only in complex one.
Easy hai......... Square me se quib ghatana hai.. It means.. Number negative hai
Thank you for such a neat solution.
6:19 nonsense. .
can be verificated complex roots to eq above ?
M^2 -M^3 = 150 ,
M= 3+√21 i,
(3+√21i)^2 - (3+√21i)^3 = 150 if we simplify this equation we will get
(a+b)^2=a^2+b^2+2ab
(a+b)^3= a^3+b^3 +3ab(a+b)
9-21+6√21i -27 +21√21i -27√21i+189 = 150 ,
198-48+27√21i-27√21i =150
150=150 , you can also check by yourself, then you will know is this right or not.
Everything went over my head 😊
No, no, it's very easy, but very monotonous and time consuming process.
Getting real nd complex numbers combination
(5.68)3 - (5.68)2. =150
The value of M= 5.68
Can we solve this equation by these step
First take m common then m(m-m^2)= 150
m=150/(m-m^2)
Then put this value and easily get m
Why should take 5 cube and 5 square
Encourage this young girl and learn from her.
Don’t give your own solutions. If you know more just stay put. Thank you friends.
Thank you for your support 😊
आपकी बात बहुत सही है।
साहिर साहब ने कहा था कि
कल और आयेंगे नगमों की खिलती कलियां चुनने वाले
मुझसे बेहतर कहने वाले तुमसे बेहतर सुनने वाले।
Giving various solutions is good , we can compare and learn all possible ways though.
M^2.(1-M)=150 --- M=-5
Am I a maths teacher but I take time to load is like I choose wrong professional
M=-5 we cah get from M^2.(1-M)=2.3.5^2 so supposing M is whole number the only possibility is M=+-5.
Apply M = 5 in the given equation, you won't get tallied on both sides.
M²-M³ = 5² - 5³ = minus 120 which is not equal to 150 as given in question
Hence 5 is wrong and minus 5 is correct answer.
M^2-M^3=150=5^2×6=5^2×(1+5)=5^2+5^3=(-5)^2-(-5)^3
M=(-5)
M^2(1-M)=150=25*6
M^2=25
1-M=6
M=1-6=-5
(-5)^2-(-5)^3=25+125=150
So m= (-5)
Complicated solution.Very easy to solve
m^2 - m^3 = 150
m^3 - m^2 + 150 = 0
(m + 5)(m^2 - 6m + 30) = 0
m = -5, 3 +/- i✓21
Step from the second to the third equation is not obvious.
M^2_3=M^_1=150
M=1/150
Easy solution:
M²-M³= 150= 5²*6
M²(1-M)= 5²*6
Now, both sides are the product of a square and a non square number
So, M²=5²
M=5
Lo hace de la forma más compleja,un problema tan sencillo
M = (-) 5 minus five.
M=-5
Hence (-5)^2 -(-5)^3
=25-(-125)
= 25+125
= 150
((-5)*2)-(-5)*3=150==> m=-5
I go to school just to be able to read and to write.
M=5
By looking at the equation one can easily say it is -5
-5 is the answer
Se ai miei tempi ci avessi così tanto e così pesantemente a risolvere questo stupido problemino il professore mi avrebbe come minimo dato solo la sufficienza
Ответ: "минус пять" находится за пять секунд
M=--5
-25-125=150
M=-5, took 5 secs
-5 answer
M= -5=(-1)5=5i^2, or, 3±√21i
Minus 5
Squares vs tribunal ,it's m
M- 30
Dear just verify yourself before posting .since d is less than 0 . It has no real root . But for we can accept it ! Anyways I enjoyed it .
M^2 -M^3 = 150 ,
M= 3+√21 i,
(3+√21i)^2 - (3+√21i)^3 = 150 if we simplify this equation we will get
(a+b)^2=a^2+b^2+2ab
(a+b)^3= a^3+b^3 +3ab(a+b)
9-21+6√21i -27 +21√21i -27√21i+189 = 150 ,
198-48+27√21i-27√21i =150
150=150 , you can also check by yourself, then you will know is this right or not.
If D less than 0 then equation has no real root.
-5=M
M=-5
-100=25-125
m^2-m^3=150
m*m*(1-m)=2*3*5*5
m*m*(m-1)=-5*5*6
m*m*(m-1)=(-5)*(-5)*(-6)
m*m*(m-1)=(-5)*(-5)*(-5-1)
m=-5 is a solution.
m^3-m^2+150=0
(m+5)*(m^2-6m+30)=0
m=-5 or m^2+6m+30=0
m=-5 or m=-3+-sqrt(3^2-30) since 9-30 is negativ, there is nooter real solution.
(M^2)^2 ➖ (m^3)^2= {m^4 ➖ m^6}= m^2 ( m ➖ 2m+2). 10^10^5^10 2^52^5^52^5 1^1^1^1^12^1 2^1 ( m ➖ 2m+1).
-5, решила устно.
In 3 sec, i have solved it, m =-5
M*2 - M*3 =150
- M*1 = 150
M = - 150
Kkkkk
-5 is obvious answer
This typ model will never ask in any compitativ xm
Bravo
no need long solution -5 is the answer
M=1/150
Aapke hisab se karne lage to zindagi me koi bhi exam crack nahi kar payenge....
25.
Nonsense done
@ satrajit ghosh 8162 is exact.
-5
M is equal to minus 5
😂EASY!!!!!!!!!!!!! M^2(1-M)=150, M
M = - 5
literally took 3 Seconds to think that answer is -5 😂 Indian Brain😂
1/150
-5
1 sec job
Solve: M^2-M^3=149
We can solve this problem by using Newton- Raphson method, follow these steps:
1. Rewrite the Equation
First, we rewrite the equation as:
f(M) = M^2 - M^3 - 149 = 0
2. Compute the Derivative
Next, compute the derivative of
f'(M) = 2M - 3M^2
3.The Newton-Raphson iteration formula is:
M_{n+1} = M_n - {f(M_n)}/{f'(M_n)}
Substitute ( f(M) ) and ( f'(M) ) into the formula:
M_{n+1} = M_n - {M_n^2 - M_n^3 - 149}/{2M_n - 3M_n^2}
4. Initial Guess
Choose an initial guess for ( M_0 ). You can start with a reasonable value like ( M_0 = 5 ), but the closer the initial guess is to the actual root, the faster the convergence
Let's perform the calculation step by step. I'll do the iteration using an initial guess of ( M_0 = 5 )
The Newton-Raphson method converged to a solution ( M approx -4.9882 ) after 12 iterations. This value of ( M ) approximately satisfies the equation ( M^2 - M^3 = 149 ).
Here are the 12 iterations of the Newton-Raphson method M_n M_{n+1}
5.0 1.1692
1.1692. -83.4847
-83.4847 -55.5533
-55.5533 -36.9416
-36.9416 -24.5543
-24.5543 -16.3416
-16.3416 -10.9663
-10.9669 -7.59547
-7.59547 -5.75294
-5.75294 -5.08055
-5.08055 -4.98978
-4.98978 -4.988210
After 12 iterations, the method converges to a solution of approximately ( M approx -4.9882 ).
M~ -4.9882
Now let's verify
M^2 - M^3 =149
(-4.9882)^2 -(-4.9882)^3 = 149
24.882 - ( -124.117) = 149
24.882+124.117= 149
148.999 = 149
149 =149
So m = -4.9882 .
@@KnowledgeTricks470 Thank you. Sincerely.
5
Устная задача.М=-5.
Nice
शेष फल प्रमेय से हल किया जा सकता है m to tha Power 2-m to tha Power 3-150=0
1.969
-5.
Mac=-5
Good
Answer: 5
150
-5 by hit and trial 😂
30 kilometres formula # confusion
0,1,-1
Complicated method.
Ans: (-5)
m={-5}
To solve for M, we'll use algebra.
Given equation: M² - M³ = 150
Rearrange: M³ - M² + 150 = 0
Factor out M²: M²(M - 1) + 150 = 0
Subtract 150 from both sides: M²(M - 1) = -150
Divide both sides by (M - 1): M² = -150 / (M - 1)
Now, solve for M:
M² = 150 / (1 - M)
M = ±√[150 / (1 - M)]
Since M must be real, let's try integer values:
M = 5 satisfies the equation.
Verify:
M² - M³ = 5² - 5³
= 25 - 125
= -100 (not 150, retry)
M = 6 satisfies the equation.
Verify:
M² - M³ = 6² - 6³
= 36 - 216
= -180 (not 150, retry)
M = 5.5 or decimal values may satisfy the equation, but let's check integers.
M = 10 satisfies the equation.
Verify:
M² - M³ = 10² - 10³
= 100 - 1000
= -900 (not 150, retry)
After retrying, we find:
M = 6.7 or M = -5.5 (approximate solutions)
However, exact integer solutions are challenging. Consider decimal approximations.
Numerical methods or graphing tools can provide precise solutions.
(-5)
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