+0

0
5
2

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
+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.

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, $$m + n + p = 8$$.

Aug 17, 2023