Separate terms with roots from terms without roots.
\[\sqrt{7-9x}=-1-3x\]
Square both sides.
\[7-9x=1+6x+9{x}^{2}\]
Move all terms to one side.
\[7-9x-1-6x-9{x}^{2}=0\]
Simplify \(7-9x-1-6x-9{x}^{2}\) to \(6-15x-9{x}^{2}\).
\[6-15x-9{x}^{2}=0\]
Factor out the common term \(3\).
\[3(2-5x-3{x}^{2})=0\]
Factor out the negative sign.
\[3\times -(3{x}^{2}+5x-2)=0\]
Divide both sides by \(3\).
\[-3{x}^{2}-5x+2=0\]
Multiply both sides by \(-1\).
\[3{x}^{2}+5x-2=0\]
Split the second term in \(3{x}^{2}+5x-2\) into two terms.
Multiply the coefficient of the first term by the constant term.
\[3\times -2=-6\]
Ask: Which two numbers add up to \(5\) and multiply to \(-6\)?
Split \(5x\) as the sum of \(6x\) and \(-x\).
\[3{x}^{2}+6x-x-2\]
\[3{x}^{2}+6x-x-2=0\]
Factor out common terms in the first two terms, then in the last two terms.
\[3x(x+2)-(x+2)=0\]
Factor out the common term \(x+2\).
\[(x+2)(3x-1)=0\]
Solve for \(x\).
Ask: When will \((x+2)(3x-1)\) equal zero?
When \(x+2=0\) or \(3x-1=0\)
Solve each of the 2 equations above.
\[x=-2,\frac{1}{3}\]
\[x=-2,\frac{1}{3}\]
Check solution
When \(x=\frac{1}{3}\), the original equation \(3x+\sqrt{7-9x}=-1\) does not hold true.We will drop \(x=\frac{1}{3}\) from the solution set.
Therefore,
x=-2
