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aev [14]
2 years ago
11

Find the zeroes of the following equation:

Mathematics
1 answer:
weqwewe [10]2 years ago
7 0

Answer:  \bold{x=\dfrac{1\pm \sqrt5}{2}}

                both zeros are irrational numbers

<u>Step-by-step explanation:</u>

Note: The question would have made more sense if if it was 2/x = x - 1 but I will answer it as written.

\dfrac{1}{x}=x-1\qquad \text{Restriction: }x\neq 0\\\\\\\text{Cross multiply, simplify, and set equal to zero:}\\1=x(x-1)\\\\1=x^2-x\\\\0=x^2-x-1\\\\\\\text{This is not factorable so you will need to use Quadratic Formula:}\\\\x=\dfrac{-(-1)\pm \sqrt{(-1)^2-4(1)(-1)}}{2(1)}\\\\\\x=\dfrac{1\pm \sqrt5}{2}

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Step-by-step explanation:

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Determine if (-2,6) and (4,12) are solutions to the system of equations:
kykrilka [37]

Answer:

Both (-2,6) and (4,12) are solutions to the system.

Step-by-step explanation:

In order to determine whether the two given points represent solutions to our system of equations, we must "plug" thos points into both equations and check that the equality remains valid.

Step 1: Plug (-2,6) into y=x+8

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The solution verifies the equation.

Step 2: Plug (-2,6) into 2y=x^2 + 8

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The solution verifies both equations. Therefore, (-2,6) is a solution to this system.

Now we must check if the second point is also valid.

Step 3: Plug (4,12) into y=x+8

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Step 4: Plug (4,12) into 2y=x^2 + 8

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The solution verifies both equations. Therefore, (4,12) is another solution to this system.

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