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sp2606 [1]
3 years ago
12

Using the quadratic formula to solve 2x2 = 4x – 7, what are the values of x? 2 plus-or-minus StartRoot 10 EndRoot i 1 plus-or-mi

nus StartRoot 10 EndRoot i StartFraction 2 plus-or-minus StartRoot 10 EndRoot i Over 2 EndFraction 1 plus-or-minus StartRoot 5 EndRoot i

Mathematics
2 answers:
grigory [225]3 years ago
8 0

Answer:

C. (Check Attached image)

Step-by-step explanation:

I'm too lasy to explain but to summarize,

- add -2x^2 to both sides

- then plug the numbers into the right places

(Google quadratic formula if you don't remember)

- solve

This post was made by the quadratic Formula gang because factoring is lame.

ioda3 years ago
7 0

Answer:

x = 1 +/- \frac{i\sqrt{10} }{2}

Step-by-step explanation:

The equation given is:

2x^2 = 4x - 7\\\\2x^2 - 4x + 7 = 0

Using quadratic formula:

x = \frac{-b +/- \sqrt{b^2 - 4ac} }{2a}

From the question:

a = 2, b = -4, c = 7

Therefore:

x = \frac{4 +/- \sqrt{(-4)^2 - 4 * 2 * 7} }{2 * 2}\\\\x = \frac{4 +/- \sqrt{(16 - 56)} }{4}\\\\x = \frac{4 +/- \sqrt{-40} }{4}\\

=> x = 1 +/- \frac{i\sqrt{10} }{2}

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Solve the following simultaneous linear congruences.
Anastaziya [24]

a. The moduli are coprime, so you can apply the Chinese remainder theorem directly. Let

x=4\cdot5+3\cdot5+3\cdot4

  • Taken mod 3, the last two terms vanish, and 20\equiv2\pmod3 so we need to multiply by the inverse of 2 modulo 3 to end up with a remainder of 1. Since 2\cdot2\equiv4\equiv1\pmod3, we multiply the first term by 2.

x=4\cdot5\cdot2+3\cdot5+3\cdot4

  • Taken mod 4, the first and last terms vanish, and 15\equiv3\pmod4. Multiply by the inverse of 3 modulo 4 (which is 3 because 3\cdot3\equiv9\equiv1\pmod4), then by 2 to ensure the proper remainder is left.

x=4\cdot5\cdot2+3\cdot5\cdot3\cdot2+3\cdot4

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x=4\cdot5\cdot2+3\cdot5\cdot3\cdot2+3\cdot4\cdot3\cdot3

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By the CRT, we have

x\equiv238\pmod{3\cdot4\cdot5}\implies x\equiv-2\pmod{60}\implies\boxed{x\equiv58\pmod{60}}

i.e. any number 58+60n (where n is an integer) satisifes the system.

b. The moduli are not coprime, so we need to check for possible contradictions. If x\equiv a\pmod m and x\equiv b\pmod n, then we need to have a\equiv b\pmod{\mathrm{gcd}(m,n)}. This basically amounts to checking that if x\equiv a\pmod m, then we should also have x\equiv a\pmod{\text{any divisor of }m}.

x\equiv4\pmod{10}\implies\begin{cases}x\equiv4\equiv0\pmod2\\x\equiv4\pmod5\end{cases}

x\equiv8\pmod{12}\implies\begin{cases}x\equiv0\pmod2\\x\equiv2\pmod3\end{cases}

x\equiv6\pmod{18}\implies\begin{cases}x\equiv0\pmod2\\x\equiv0\pmod3\end{cases}

The last congruence conflicts with the previous one modulo 3, so there is no solution to this system.

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Answer:

5, assuming you get the worst possible slips off: 0,1,2,3,4

Step-by-step explanation:

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