Question

question
solve for all possible values of x.
\\(\sqrt{44 - 5x} = x - 6\\)
answer attempt 1 out of 2
\\(\oplus\\) additional solution \\(\circleddash\\) no solution
\\(x = \square\\) submit answer

Explanation:

Step1: Square both sides to eliminate the square root

To solve the equation \(\sqrt{44 - 5x}=x - 6\), we first square both sides. Squaring the left side \((\sqrt{44 - 5x})^2\) gives \(44 - 5x\), and squaring the right side \((x - 6)^2\) using the formula \((a - b)^2=a^2-2ab + b^2\) (here \(a = x\) and \(b = 6\)) gives \(x^{2}-12x + 36\). So we get the equation:
\(44-5x=x^{2}-12x + 36\)

Step2: Rearrange into standard quadratic form

Move all terms to one side to form a quadratic equation. Subtract \(44\) and add \(5x\) to both sides:
\(x^{2}-12x + 36-44 + 5x=0\)
Combine like terms: \(x^{2}-7x - 8 = 0\)

Step3: Factor the quadratic equation

We need to find two numbers that multiply to \(-8\) and add to \(-7\). The numbers are \(-8\) and \(1\). So we can factor the quadratic as:
\((x - 8)(x+ 1)=0\)

Step4: Solve for x

Using the zero - product property, if \(ab = 0\), then either \(a = 0\) or \(b = 0\). So we set each factor equal to zero:

• For \(x - 8=0\), we get \(x = 8\).
• For \(x + 1=0\), we get \(x=-1\).

Step5: Check for extraneous solutions

We need to check these solutions in the original equation because squaring both sides can introduce extraneous solutions.

• For \(x = 8\):

Left side: \(\sqrt{44-5\times8}=\sqrt{44 - 40}=\sqrt{4}=2\)
Right side: \(8 - 6=2\)
Since left side = right side, \(x = 8\) is a valid solution.

• For \(x=-1\):

Left side: \(\sqrt{44-5\times(-1)}=\sqrt{44 + 5}=\sqrt{49}=7\)
Right side: \(-1-6=-7\)
Since \(7
eq - 7\), \(x=-1\) is an extraneous solution.

Answer:

\(x = 8\)