We have an answer to "Solving 3x2+4x−2=0 (mod 31)"

High School
Algebra
Algebra I
Quadratic function and equation

monkeyman130yb

Answered question

2022-04-03

Solving

3 x^{2} + 4 x - 2 = 0

(mod 31)

Answer & Explanation

pastuh7vka

Beginner2022-04-04Added 13 answers

Step 1
Might be easier to factor or use the quadratic formula.
$3 x^{2} + 4 x - 2 \equiv 0 \pm \mod 31$
so abusing notation where $\sqrt{k}$ will mean the congruence a where $a^{2} \equiv k \pm \mod 31$ and $\frac{1}{m} = m^{- 1}$ is the congruence where $m (m^{- 1}) \equiv 1 \pm \mod 31$ then
$x \equiv \frac{- 4 \pm \sqrt{16 + 24}}{6} \equiv$
$(- 4 \pm \sqrt{40}) \cdot 6^{- 1} \equiv$
$(27 \pm \sqrt{9}) \cdot 6^{- 1} \equiv$
$(27 \pm 3) \cdot 6^{- 1} \equiv$
${\begin{cases} 30 \cdot 6^{- 1} \equiv 5 \cdot 6 \cdot 6^{- 1} \equiv 5 \\ 24 \cdot 6^{- 1} \equiv 4 \cdot 6 \cdot 6^{- 1} \equiv 4 \end{cases} (\mod 31)$
So $x \equiv 4 \pm \mod 31$ or $x \equiv 5 \pm \mod 31$

microsgopx6z7

Beginner2022-04-05Added 14 answers

Step 1
Completing the square, we have:

5 x^{2} \equiv 2 {(x - 1)}^{2}

Since

36 \equiv 31 + 5 \equiv 5

, we have:

36 x^{2} \equiv 2 {(x - 1)}^{2}

18 x^{2} \equiv {(x - 1)}^{2}

Adding 31 again:

49 x^{2} \equiv {(x - 1)}^{2}

Then

7 x \equiv x - 1

- 7 x \equiv x - 1

Which is the same as

6 x \equiv - 1 \equiv 30

8 x \equiv 1 \equiv 32

Giving us the final answer of

x \equiv 4, 5

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