Question

solve for x, where x is a real number.\\(\sqrt{-x + 30} - \sqrt{x + 31} = -1\\)\\(\\) (if there is more than one solution, separate them with commas.)\\(x = \square\\)

Explanation:

Step1: Isolate one square root

Start with the equation \(\sqrt{-x + 30}-\sqrt{x + 31}=-1\). Add \(\sqrt{x + 31}\) to both sides:
\(\sqrt{-x + 30}=\sqrt{x + 31}-1\)

Step2: Square both sides

Square both sides to eliminate the square root on the left. Recall that \((a - b)^2=a^2-2ab + b^2\), so:
\((\sqrt{-x + 30})^2=(\sqrt{x + 31}-1)^2\)
\(-x + 30=(x + 31)-2\sqrt{x + 31}+1\)
Simplify the right - hand side:
\(-x + 30=x + 32-2\sqrt{x + 31}\)

Step3: Isolate the remaining square root

Subtract \(x + 32\) from both sides:
\(-x + 30-(x + 32)=-2\sqrt{x + 31}\)
\(-x+30 - x - 32=-2\sqrt{x + 31}\)
\(-2x - 2=-2\sqrt{x + 31}\)
Divide both sides by \(- 2\):
\(x + 1=\sqrt{x + 31}\)

Step4: Square both sides again

Square both sides to eliminate the square root:
\((x + 1)^2=(\sqrt{x + 31})^2\)
\(x^{2}+2x + 1=x + 31\)

Step5: Solve the quadratic equation

Rearrange the equation to standard quadratic form \(ax^{2}+bx + c = 0\):
\(x^{2}+2x + 1-x - 31 = 0\)
\(x^{2}+x - 30 = 0\)
Factor the quadratic. We need two numbers that multiply to \(-30\) and add to \(1\). The numbers are \(6\) and \(-5\):
\((x + 6)(x - 5)=0\)
Set each factor equal to zero:
\(x+6 = 0\) or \(x - 5=0\)
So \(x=-6\) or \(x = 5\)

Step6: Check for extraneous solutions

• For \(x=-6\):

Left - hand side of the original equation: \(\sqrt{-(-6)+30}-\sqrt{-6 + 31}=\sqrt{36}-\sqrt{25}=6 - 5 = 1\)
Right - hand side: \(-1\)
Since \(1
eq-1\), \(x=-6\) is an extraneous solution.

• For \(x = 5\):

Left - hand side of the original equation: \(\sqrt{-5 + 30}-\sqrt{5 + 31}=\sqrt{25}-\sqrt{36}=5 - 6=-1\)
Right - hand side: \(-1\)
Since \(-1=-1\), \(x = 5\) is a valid solution.

Answer:

\(5\)