Fastest Method | Solving This Mathematics Problem | Math Olympiad Problem.

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OnlineMaths TV

OnlineMaths TV

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@henrya.bbruce4854
@henrya.bbruce4854 16 сағат бұрын
Am a chemist but after following you , am now more mathematician than a chemist.
@LeonardoAntónioTchipuapuaCalem
@LeonardoAntónioTchipuapuaCalem 5 күн бұрын
Correct
@ElvisSaturn
@ElvisSaturn 5 күн бұрын
√m +√-m = 12 -> √m(1+i) = 12 -> √m = 12/(1+i) -> m = ±12²/(1+i)² -> m = ±144/(1-1+2i) -> m=±72/i = ∓72i (Fastest Method )
@ABHISHEKKUMAR-01024
@ABHISHEKKUMAR-01024 4 күн бұрын
Arbitrary approach: Given equation is √m + √( - m) = 12 ...(1) Let √m = p and √( - m) = q Then m = p² and - m = q² Then p² + q² = 0 ...(2) Also equation (1) becomes p + q = 12 ...(3) Now, (p + q)² = p² + 2pq + q² = (p² + q²) + 2pq Putting p + q = 12 and p² + q² = 0, we get (12)² = 0 + 2pq or, 144 = 2pq or, 72 = pq ...(4) Now, (p - q)² = (p + q)² - 4pq = (12)² - 4(72) = 144 - 288 = - 144 = 144i² = (12i)² => p - q = ± 12i Now, 2p = (p + q) + (p - q) = 12 + ( ± 12i ) = 12 + 12i, 12 - 12i => p = 6 + 6i, 6 - 6i => √m = 6 + 6i, 6 - 6i => m = (6 + 6i)², (6 - 6i)² = 6² + 2(6)(6i) + (6i)², 6² - 2(6)(6i) + (6i)² = 36 + 72i + 36i², 36 - 72i + 36i² = 36 + 72i + ( - 36 ) , 36 - 72i + ( - 36 ) = 72i, - 72i = ± 72i Another approach : Let √m = p and √( - m) = q Then p + q = 12 ...(1) Also, m = p² and - m = q² Now, p² - q² = m - ( - m) or, p² - q² = 2m or, (p + q)(p - q) = 2m or, 12 (p - q) = 2m or, p - q = m/6 Now, 2p = (p + q) + (p - q) or, 2p = 12 + (m/6) or, 2√m = 12 + (m/6) or, (2√m)² = { 12 + (m/6) }² or, 4m = (12)² + 2(12)(m/6) + (m/6)² or, 4m = 144 + (24)(m/6) + (m²/36) or, 4m = 144 + 4m + (m²/36) or, 0 = 144 + (m²/36) or, 144 + (m²/36) = 0 or, (m²/36) = - 144 or, m² = - 36 × 144 or, m² = - 6² × 12² or, m² = - (6 × 12)² or, m² = - 72² or, m² = 72²i² or, m² = (72i)² or, m = ± 72i And many more approaches...
@wes9627
@wes9627 5 күн бұрын
Let m=±ai where a>0 is real. Then m and -m lie on opposite sides of a radius a circle centered on the origin of the complex plane. Now √(±m) = √a√(±i)=√a[cos(±π/4)+i*sin(±π/4)] Then √m+√(-m)=√a[cos(π/4)+i*sin(π/4)]+√a[cos(-π/4)+i*sin(±π/4)]=2√a(√2/2)=12 It follows that √a=12/√2, a=144/2=72, and m=±72i.
@RyanLewis-Johnson-wq6xs
@RyanLewis-Johnson-wq6xs 5 күн бұрын
Sqrt[72i]+Sqrt[-72i] m=±72i It’s in my head.
@onlineMathsTV
@onlineMathsTV 5 күн бұрын
Nice one, that is while you are the best sir. Maximum respect sir.
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