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I need a check of this answer

 

The roots of $x(3x-7)=-3+(x-1)(x-4)$ can be expressed in the form $\frac{m+\sqrt{n}}{p}$ and $\frac{m-\sqrt{n}}{p}$, where $m$, $n$, and $p$ are positive integers that have a greatest common divisor of $1$.  Find $m+n+p$.

 Aug 17, 2023
 #1
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No you don't you want us to do it for you. You try to give the illusion that you have solved it while simply copy-pasting your stupid homework question and giving us no information on your retarded thought-process whatsoever.

 Aug 17, 2023
edited by Guest  Aug 17, 2023
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avatar+121 
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Certainly, let's solve the given equation step by step:

Starting with the equation:
\[x(3x - 7) = -3 + (x - 1)(x - 4)\]

Expand the right side:
\[x(3x - 7) = -3 + x^2 - 5x + 4\]

Simplify:
\[3x^2 - 7x = x^2 - 5x + 1\]

Subtract \(x^2\) and \(5x\) from both sides:
\[2x^2 - 2x = 1\]

Divide both sides by 2:
\[x^2 - x = \frac{1}{2}\]

Complete the square on the left side:
\[x^2 - x + \left(\frac{1}{2}\right)^2 = \frac{1}{2} + \left(\frac{1}{2}\right)^2\]

Simplify:
\[x^2 - x + \frac{1}{4} = \frac{5}{4}\]

Factor the left side:
\[\left(x - \frac{1}{2}\right)^2 = \frac{5}{4}\]

Take the square root of both sides:
\[x - \frac{1}{2} = \pm \frac{\sqrt{5}}{2}\]

Solve for \(x\):
\[x = \frac{1 \pm \sqrt{5}}{2}\]

So, the roots of the equation are:
\[x = \frac{1 + \sqrt{5}}{2} \quad \text{and} \quad x = \frac{1 - \sqrt{5}}{2}\]

Now let's write these roots in the given form:
\[x = \frac{1 + \sqrt{5}}{2} = \frac{1 + \sqrt{5}}{2} \cdot \frac{2}{2} = \frac{2 + \sqrt{5}}{2} = \frac{2}{2} + \frac{\sqrt{5}}{2} = 1 + \frac{\sqrt{5}}{2}\]

Similarly,
\[x = \frac{1 - \sqrt{5}}{2} = 1 - \frac{\sqrt{5}}{2}\]

Now we can find \(m + n + p\):
\[m + n + p = 1 + 5 + 2 = 8\]

Therefore, \(m + n + p = 8\).

 Aug 17, 2023

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