Solve for \(x\) in \((x+1)\sqrt{{x}^{2}+x-2}=2x+2\).
Solve for \(x\).
\[(x+1)\sqrt{{x}^{2}+x-2}=2x+2\]
Factor \({x}^{2}+x-2\).
Ask: Which two numbers add up to \(1\) and multiply to \(-2\)?
Rewrite the expression using the above.
\[(x-1)(x+2)\]
\[(x+1)\sqrt{(x-1)(x+2)}=2x+2\]
Square both sides.
\[{(x+1)}^{2}(x-1)(x+2)=4{x}^{2}+8x+4\]
Expand.
\[{x}^{4}+2{x}^{3}-{x}^{3}-2{x}^{2}+2{x}^{3}+4{x}^{2}-2{x}^{2}-4x+{x}^{2}+2x-x-2=4{x}^{2}+8x+4\]
Simplify \({x}^{4}+2{x}^{3}-{x}^{3}-2{x}^{2}+2{x}^{3}+4{x}^{2}-2{x}^{2}-4x+{x}^{2}+2x-x-2\) to \({x}^{4}+3{x}^{3}+{x}^{2}-3x-2\).
\[{x}^{4}+3{x}^{3}+{x}^{2}-3x-2=4{x}^{2}+8x+4\]
Move all terms to one side.
\[{x}^{4}+3{x}^{3}+{x}^{2}-3x-2-4{x}^{2}-8x-4=0\]
Simplify \({x}^{4}+3{x}^{3}+{x}^{2}-3x-2-4{x}^{2}-8x-4\) to \({x}^{4}+3{x}^{3}-3{x}^{2}-11x-6\).
\[{x}^{4}+3{x}^{3}-3{x}^{2}-11x-6=0\]
Factor \({x}^{4}+3{x}^{3}-3{x}^{2}-11x-6\) using Polynomial Division.
Factor the following.
\[{x}^{4}+3{x}^{3}-3{x}^{2}-11x-6\]
First, find all factors of the constant term 6.
\[1, 2, 3, 6\]
Try each factor above using the Remainder Theorem.
Substitute 1 into x. Since the result is not 0, x-1 is not a factor..\({1}^{4}+3\times {1}^{3}-3\times {1}^{2}-11\times 1-6 = -16\)
Substitute -1 into x. Since the result is 0, x+1 is a factor..\({(-1)}^{4}+3{(-1)}^{3}-3{(-1)}^{2}-11\times -1-6 = 0\)
\[x+1\]
Polynomial Division: Divide \({x}^{4}+3{x}^{3}-3{x}^{2}-11x-6\) by \(x+1\).
| \[x^3\] | \[2x^2\] | \[-5x\] | \[-6\] | |
| \[x+1\] | \[x^4\] | \[3x^3\] | \[-3x^2\] | \[-11x\] | \[-6\] |
| \[x^4\] | \[x^3\] | | | |
| | \[2x^3\] | \[-3x^2\] | \[-11x\] | \[-6\] |
| | \[2x^3\] | \[2x^2\] | | |
| | | \[-5x^2\] | \[-11x\] | \[-6\] |
| | | \[-5x^2\] | \[-5x\] | |
| | | | \[-6x\] | \[-6\] |
| | | | \[-6x\] | \[-6\] |
| | | | | \[\] |
Rewrite the expression using the above.
\[{x}^{3}+2{x}^{2}-5x-6\]
\[({x}^{3}+2{x}^{2}-5x-6)(x+1)=0\]
Factor \({x}^{3}+2{x}^{2}-5x-6\) using Polynomial Division.
Factor the following.
\[{x}^{3}+2{x}^{2}-5x-6\]
First, find all factors of the constant term 6.
\[1, 2, 3, 6\]
Try each factor above using the Remainder Theorem.
Substitute 1 into x. Since the result is not 0, x-1 is not a factor..\({1}^{3}+2\times {1}^{2}-5\times 1-6 = -8\)
Substitute -1 into x. Since the result is 0, x+1 is a factor..\({(-1)}^{3}+2{(-1)}^{2}-5\times -1-6 = 0\)
\[x+1\]
Polynomial Division: Divide \({x}^{3}+2{x}^{2}-5x-6\) by \(x+1\).
| \[x^2\] | \[x\] | \[-6\] | |
| \[x+1\] | \[x^3\] | \[2x^2\] | \[-5x\] | \[-6\] |
| \[x^3\] | \[x^2\] | | |
| | \[x^2\] | \[-5x\] | \[-6\] |
| | \[x^2\] | \[x\] | |
| | | \[-6x\] | \[-6\] |
| | | \[-6x\] | \[-6\] |
| | | | \[\] |
Rewrite the expression using the above.
\[{x}^{2}+x-6\]
\[({x}^{2}+x-6)(x+1)(x+1)=0\]
Factor \({x}^{2}+x-6\).
Ask: Which two numbers add up to \(1\) and multiply to \(-6\)?
Rewrite the expression using the above.
\[(x-2)(x+3)\]
\[(x-2)(x+3)(x+1)(x+1)=0\]
Solve for \(x\).
Ask: When will \((x-2)(x+3)(x+1)(x+1)\) equal zero?
When \(x-2=0\), \(x+3=0\), \(x+1=0\), or \(x+1=0\)
Solve each of the 4 equations above.
\[x=2,-3,-1\]
\[x=2,-3,-1\]
Check solution
When \(x=-1\), the original equation \((x+1)\sqrt{{x}^{2}+x-2}=2x+2\) does not hold true.We will drop \(x=-1\) from the solution set.
Therefore,
Substitute \(x=2,-3\) into \(\sqrt{x+5}-\sqrt{x-3}=2\).
Start with the original equation.
\[\sqrt{x+5}-\sqrt{x-3}=2\]
Let \(x=2,-3\).
\[\sqrt{(2,-3)+5}-\sqrt{(2,-3)-3}=2\]
Since \(\sqrt{(2,-3)+5}-\sqrt{(2,-3)-3}=2\) is not true, this is an inconsistent system.
