Question

Solve the equation (equation will be posted in the comments section)

Answer 1

-3=\[\sqrt{22-x}\]-\[\sqrt{3x-3}\]

Answer 2

lmk if that works

Answer 3

\[\sqrt{22-x}-\sqrt{3x-3}=-3\]
\[let ~a=\sqrt{22-x},b=\sqrt{3x-3}\]
a-b=-3 ...(1)
\[a^2-b^2=22-x-(3x-3)=22-x-3x+3=25-4x\]
\[\frac{ a^2-b^2 }{ a-b }=\frac{ 25-4x }{ -3 }\]
\[a+b=\frac{ 25-4x }{ -3 } ...(2)\]
\[2a=\frac{ 25-4x }{ -3 }-3=\frac{ 25-4x+9 }{ -3 }=\frac{ 34-4x }{ -3 }\]
\[a=\frac{ 17-2x }{ -3 }=\frac{ 2x-17 }{ 3 }\]
\[\sqrt{22-x}=\frac{ 2x-17 }{ 3 }\]
squaring
\[9(22-x)=(2x-17)^2\]
\[198-9x=4x^2+289-68x\]
\[4x^2-68x+289+9x-198=0\]
\[4x^2-59x+91=0\]
\[x=\frac{ 59 \pm \sqrt{59^2-4 *4*91} }{2*4 }\]
\[x=\frac{ 59 \pm \sqrt{3481-1456} }{ 8 }\]
\[x=\frac{ 59 \pm \sqrt{2025} }{ 8}\]
\[x=\frac{ 59 \pm 45 }{ 8 }\]
\[either ~x=\frac{ 59+45 }{ 8 }=\frac{ 104 }{ 8}=13\]
\[or ~x=\frac{ 59-45 }{ 8 }=\frac{ 14 }{ 8}=\frac{ 7 }{ 4}\]
x=7/4 does not satisfy the equation.
so it is an exteraneous solution.
Hence the solution is x=13

Answer 4

or you can solve it this way.
\[\sqrt{22-x}=-3+\sqrt{3x-3}\]
squaring
\[22-x=9+3x-3-6\sqrt{3x-3}\]
\[6\sqrt{3x-3}=3x+6+x-22\]
\[6\sqrt{3x-3}=4x-16\]
divide by 2
\[3\sqrt{3x-3}=2x-8\]
again squaring
\[9(3x-3)=4x^2+64-32x\]
\[27x-27=4x^2+64-32x\]
\[4x^2-32x-27x+64+27\]
\[4x^2-59x+91=0\]
further i have solved above.