Question

question
solve for all possible values of x.
\\(\sqrt{8x + 41} = x + 6\\)

Explanation:

Step1: Square both sides

To eliminate the square root, square both sides of the equation \(\sqrt{8x + 41}=x + 6\). This gives us \((\sqrt{8x + 41})^2=(x + 6)^2\), which simplifies to \(8x + 41=x^2+12x + 36\).

Step2: Rearrange into quadratic form

Rearrange the equation \(8x + 41=x^2+12x + 36\) to standard quadratic form \(ax^2+bx + c = 0\). Subtract \(8x\) and \(41\) from both sides: \(x^2+12x+36 - 8x - 41 = 0\), which simplifies to \(x^2 + 4x-5=0\).

Step3: Factor the quadratic

Factor the quadratic equation \(x^2 + 4x - 5 = 0\). We need two numbers that multiply to \(-5\) and add to \(4\). These numbers are \(5\) and \(-1\), so the equation factors as \((x + 5)(x - 1)=0\).

Step4: Solve for x

Set each factor equal to zero and solve for \(x\):

• For \(x + 5=0\), we get \(x=-5\).
• For \(x - 1=0\), we get \(x = 1\).

Step5: Check solutions

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

• For \(x=-5\): Substitute into \(\sqrt{8x + 41}\) and \(x + 6\). \(\sqrt{8(-5)+41}=\sqrt{-40 + 41}=\sqrt{1}=1\), and \(x + 6=-5 + 6 = 1\). So \(x=-5\) is a valid solution.
• For \(x = 1\): Substitute into \(\sqrt{8x + 41}\) and \(x + 6\). \(\sqrt{8(1)+41}=\sqrt{8 + 41}=\sqrt{49}=7\), and \(x + 6=1 + 6 = 7\). So \(x = 1\) is a valid solution.

Answer:

The possible values of \(x\) are \(x=-5\) and \(x = 1\)