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x+1=\sqrt{5x+11}
Subtract -1 from both sides of the equation.
\left(x+1\right)^{2}=\left(\sqrt{5x+11}\right)^{2}
Square both sides of the equation.
x^{2}+2x+1=\left(\sqrt{5x+11}\right)^{2}
Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+1\right)^{2}.
x^{2}+2x+1=5x+11
Calculate \sqrt{5x+11} to the power of 2 and get 5x+11.
x^{2}+2x+1-5x=11
Subtract 5x from both sides.
x^{2}-3x+1=11
Combine 2x and -5x to get -3x.
x^{2}-3x+1-11=0
Subtract 11 from both sides.
x^{2}-3x-10=0
Subtract 11 from 1 to get -10.
a+b=-3 ab=-10
To solve the equation, factor x^{2}-3x-10 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.
1,-10 2,-5
Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -10.
1-10=-9 2-5=-3
Calculate the sum for each pair.
a=-5 b=2
The solution is the pair that gives sum -3.
\left(x-5\right)\left(x+2\right)
Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.
x=5 x=-2
To find equation solutions, solve x-5=0 and x+2=0.
5=\sqrt{5\times 5+11}-1
Substitute 5 for x in the equation x=\sqrt{5x+11}-1.
5=5
Simplify. The value x=5 satisfies the equation.
-2=\sqrt{5\left(-2\right)+11}-1
Substitute -2 for x in the equation x=\sqrt{5x+11}-1.
-2=0
Simplify. The value x=-2 does not satisfy the equation.
x=5
Equation x+1=\sqrt{5x+11} has a unique solution.