A Quadratic Equation | Problem 363

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Опубликовано:

24 сен 2024

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Комментарии : 7

@scottleung9587 8 дней назад

Got 'em both!

@mcwulf25 8 дней назад

Good problem. I was mentally shouting "16 and 25" ehen you were solving for b 😅

@quneptune 8 дней назад

funfact: a^1/x =(ˣ√a)e^(2inπ/x) and the symbol ± can always be replaced with e^inπ

@aekben7312 8 дней назад

I want to emphasize how to find a square root a complex number: Let D:Delta a descriminent of quadratic equation D=6^2-4×1×(-40) D=36+160i Let (D)^1/2=x+iy Square both sides D=(x+iy)^2=36+160i ~ x^2-y^2+2xy=36+160i ~{x^2-y^2=36,2xy=160}~ {x^-y^2=36...(1) xy=80.... (2)} In same time The modulus of D(Delta)= x^2+y^2=36^2+160^2~ x^2+y^2=164...Eq(3) Then the system of becomes: { x^2-y^2=36..... (1) x^2+y^2=164....(3) xy=80........... (2)} (1)+(3)~2x^2=36+164 x^2=100 x=10,x=-10 (3) -(1)~2y^2=164-36~ y^2=64 y=8,y=-8 Let (Delta)^1/2=10+8i Z1={-6-(10+8i)}÷2 Z1=-8-4i Z2={-6+(10+8i)}÷2 Z2=2+4i

@aplusbi 7 дней назад

This is amazing! Thanks

@chaosredefined3834 7 дней назад

Let a and b be the real values such that z = a + bi. Therefore: z^2 = a^2 - b^2 + 2abi z^2 + 6z = a^2 - b^2 + 6a + (2ab + 6b)i = 40i So, a^2 - b^2 + 6a = 0, and 2ab + 6b = 40 a^2 + 6a = b^2 a^2 + 6a + 9 = b^2 + 9 (a+3)^2 = b^2 + 9 2ab + 6b = 40 2b(a + 3) = 40 4b^2 (a + 3)^2 = 1600 4b^2 (b^2 + 9) = 1600 Let c = b^2 4c(c + 9) = 1600 4c^2 + 36c = 1600 Note that (2c + 9)^2 = 4c^2 + 36c + 81 4c^2 + 36c + 81 = 1681 (2c + 9)^2 = 41^2 2c + 9 = 41 or 2c + 9 = -41 Note that c = b^2, so c > 0. Therefore, 2c > 0 and 2c + 9 > 9. Hence, 2c + 9 =/= -41 2c + 9 = 41 2c = 32 c = 16 Since c = b^2, then b = 4 or -4. And since (a + 3)^2 = b^2 + 9, (a + 3)^2 = 16 + 9 = 25, so a+3 = 5 or a+3 = -5. This means that a = 2 or a = -8 From the statement that 2b(a + 3) = 40, we get that either a = 2, b = 4, or a = -8, b = -4. So, the solutions are z = 2 + 4i or z = -8 - 4i.

@aplusbi 7 дней назад

Good thinking!