2x 2 X 6 0

5 Answers
5

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Since $53$ is prime, it divides either $x-2$ or $2x+3$. Thus $x\equiv 2,\,25\,(\operatorname{mod}53)$.

answered
May 13, 2022 at 9:52

J.G.'s user avatar

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    If we put another number instead of $53$ , for example $54$ , what we can do then ?
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    May 13, 2022 at 9:54

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    @S.H.W. That’s tricky because for composite numbers there are a number of factorisations that would work. In general, however, there will be a set of modulo classes of $x$ that comprise the solution, so you could always just go through all of them.
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    May 13, 2022 at 9:57

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    @S.H.W That’s more complex, since the ring $\mathbb{Z}_{54}$ has zero divisors.
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    May 13, 2022 at 9:59

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    Okay , thanks . Do you know any geometric solution for this kelainan ?
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    May 13, 2022 at 10:03

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    @S.H.W What do you mean by geometric? BTW one thing you could do is write $54$ as a product of prime powers ($2\times 27$), then solve simultaneous equations. Since $x-2,\,2x+3$ can’t both be divisible by $3$, one will in fact be divisible by $27$.
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    May 13, 2022 at 10:06

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It follows from your computations that $x=2$ will work.

If you’re after
all
solutions (modulo $53$), you do\begin{align}(2x+3)(x-2)\equiv0\pmod{53}&\iff 2x+3\equiv0\pmod{53}\vee x-2\equiv0\pmod{53}\\&\iff2x\equiv50\pmod{53}\vee x\equiv2\pmod{53}\\&\iff x\equiv25\pmod{53}\vee x\equiv2\pmod{53}.\end{align}

answered
May 13, 2022 at 9:54

José Carlos Santos's user avatar

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    If we put another number instead of $53$ , for example $54$ , what we can do then ?
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    May 13, 2022 at 9:56

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See Euclid’s Lemma. $53$ divides the product $(2x+3)(x-2)$. Then $53$ divides one of them, or both. You now have all the cases.

answered
May 13, 2022 at 9:55

poyea's user avatar

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You can do also like this. Multiply this with $8$ $$2x^2 – x – 6 \equiv 0 \mod 53$$ so $$16x^2 – 8x – 48 \equiv 0 \mod 53$$ so $$16x^2 – 8x +1- 49 \equiv 0 \mod 53$$ so $$(4x-1)^2 \equiv 7^2 \mod 53$$ so $$4x-1 \equiv \pm 7 \mod 53$$ …


Note
that if you have $$ax^2+bx+c\equiv 0 \mod d$$ and $\gcd(2a,d) =1$ then you can do this always (multiply with $4a$): $$4a^2x^2+4abx+4ac\equiv 0 \mod d$$ so $$4a^2x^2+4abx+b^2\equiv b^2-4ac \mod d$$ $$(2ax+b)^2\equiv \Delta \mod d$$ where $\Delta = b^2-4ac$.

answered
May 13, 2022 at 10:05

nonuser's user avatar

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$53=1*53$(it is a prime number)

So once let, $2x+3=53$
And second time let $x-2=53$
For (i) $x=25$
$\Rightarrow$ $Number=53*27$

Similarly for (ii) $x=55$
$\Rightarrow$ $Number=53*113$

Now considering another case when $53|2x+3$ or $53|x-2$
Which means (i) $53n=2x+3$ (Here cakrawala can take any odd integral value)
OR
(ii) $53n=x-2$(here n can take any integral value)

It means there can’lengkung langit be any discrete value/s of x

answered
May 13, 2022 at 9:55

Love Invariants's user avatar

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Source: https://math.stackexchange.com/questions/2779059/determine-x-such-that-2×2-x-6-be-a-multiple-of-53