Calculate $\sqrt{x-1}$ to the power of $2$ and get $x-1$.
$$9\left(x-1\right)=\left(-1-x\right)^{2}$$
Use the distributive property to multiply $9$ by $x-1$.
$$9x-9=\left(-1-x\right)^{2}$$
Use binomial theorem $\left(a-b\right)^{2}=a^{2}-2ab+b^{2}$ to expand $\left(-1-x\right)^{2}$.
$$9x-9=1+2x+x^{2}$$
Subtract $2x$ from both sides.
$$9x-9-2x=1+x^{2}$$
Combine $9x$ and $-2x$ to get $7x$.
$$7x-9=1+x^{2}$$
Subtract $x^{2}$ from both sides.
$$7x-9-x^{2}=1$$
Subtract $1$ from both sides.
$$7x-9-x^{2}-1=0$$
Subtract $1$ from $-9$ to get $-10$.
$$7x-10-x^{2}=0$$
Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.
$$-x^{2}+7x-10=0$$
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as $-x^{2}+ax+bx-10$. To find $a$ and $b$, set up a system to be solved.
$$a+b=7$$ $$ab=-\left(-10\right)=10$$
Since $ab$ is positive, $a$ and $b$ have the same sign. Since $a+b$ is positive, $a$ and $b$ are both positive. List all such integer pairs that give product $10$.
$$1,10$$ $$2,5$$
Calculate the sum for each pair.
$$1+10=11$$ $$2+5=7$$
The solution is the pair that gives sum $7$.
$$a=5$$ $$b=2$$
Rewrite $-x^{2}+7x-10$ as $\left(-x^{2}+5x\right)+\left(2x-10\right)$.
$$\left(-x^{2}+5x\right)+\left(2x-10\right)$$
Factor out $-x$ in the first and $2$ in the second group.
$$-x\left(x-5\right)+2\left(x-5\right)$$
Factor out common term $x-5$ by using distributive property.
$$\left(x-5\right)\left(-x+2\right)$$
To find equation solutions, solve $x-5=0$ and $-x+2=0$.
$$x=5$$ $$x=2$$
Substitute $5$ for $x$ in the equation $x-3\sqrt{x-1}+1=0$.
$$5-3\sqrt{5-1}+1=0$$
Simplify. The value $x=5$ satisfies the equation.
$$0=0$$
Substitute $2$ for $x$ in the equation $x-3\sqrt{x-1}+1=0$.
