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 $m+n+p$.

sandwich Aug 17, 2023

#1**+1 **

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.

Guest Aug 17, 2023

edited by
Guest
Aug 17, 2023

#2**-1 **

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

SpectraSynth Aug 17, 2023