Question

solve for x, where x is a real number.\
\\(\sqrt{8x + 9} = x\\)\
if there is more than one solution, separate them with commas.\
if there is no solution, click on
o solution\.\
x = \\(\quad\\)\
no solution, \\(\sqrt{\square}\\), \\(\sqrt\square{\square}\\), \\(\frac{\square}{\square}\\), \\(\square\square\dots\\)

Explanation:

Step1: Square both sides to eliminate the square root

To solve the equation \(\sqrt{8x + 9}=x\), we first square both sides of the equation. Squaring the left side \(\sqrt{8x + 9}\) gives us \(8x + 9\), and squaring the right side \(x\) gives us \(x^{2}\). So we get the quadratic equation:

$$x^{2}=8x + 9$$

Step2: Rearrange into standard quadratic form

We rearrange the equation \(x^{2}=8x + 9\) to the standard quadratic form \(ax^{2}+bx + c = 0\) by subtracting \(8x\) and \(9\) from both sides:

$$x^{2}-8x - 9=0$$

Step3: Factor the quadratic equation

We factor the quadratic equation \(x^{2}-8x - 9 = 0\). We need two numbers that multiply to \(- 9\) and add up to \(-8\). The numbers are \(-9\) and \(1\). So we can factor the equation as:

$$(x - 9)(x+ 1)=0$$

Step4: Solve for x using the zero - product property

Using the zero - product property, if \((x - 9)(x + 1)=0\), then either \(x-9=0\) or \(x + 1=0\).

• If \(x-9=0\), then \(x=9\).
• If \(x + 1=0\), then \(x=-1\).

Step5: Check for extraneous solutions

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

• For \(x = 9\):

Substitute \(x = 9\) into the original equation \(\sqrt{8x+9}\). We have \(\sqrt{8\times9 + 9}=\sqrt{72 + 9}=\sqrt{81}=9\), and the right side of the original equation is \(x = 9\). So \(x = 9\) is a valid solution.

• For \(x=-1\):

Substitute \(x=-1\) into the original equation \(\sqrt{8x + 9}\). We have \(\sqrt{8\times(-1)+9}=\sqrt{-8 + 9}=\sqrt{1}=1\), and the right side of the original equation is \(x=-1\). Since \(1
eq - 1\), \(x=-1\) is an extraneous solution.

Answer:

\(9\)