Question

solve for ( x );
( x - 3 = sqrt{4x - 7} )
(dont forget to check for extraneous solutions!)
( \bigcirc ) -8, -2
( \bigcirc ) 2
( \bigcirc ) 8, 2
( \bigcirc ) 8
( \bigcirc ) -8

Explanation:

Step1: Square both sides to eliminate the square root

To solve the equation \( x - 3=\sqrt{4x - 7} \), we first square both sides. Squaring the left side \((x - 3)^2\) gives \(x^{2}-6x + 9\), and squaring the right side \((\sqrt{4x - 7})^2\) gives \(4x-7\). So we get the quadratic equation:
\(x^{2}-6x + 9=4x-7\)

Step2: Rearrange into standard quadratic form

Subtract \(4x\) and add \(7\) to both sides to get the equation in standard quadratic form \(ax^{2}+bx + c = 0\).
\(x^{2}-6x-4x + 9 + 7=0\)
\(x^{2}-10x + 16=0\)

Step3: Factor the quadratic equation

We factor the quadratic \(x^{2}-10x + 16\). We need two numbers that multiply to \(16\) and add up to \(- 10\). The numbers are \(-8\) and \(-2\). So,
\((x - 8)(x - 2)=0\)

Step4: Solve for x

Using the zero - product property, if \((x - 8)(x - 2)=0\), then either \(x-8 = 0\) or \(x - 2=0\).

• If \(x-8=0\), then \(x = 8\).
• If \(x - 2=0\), then \(x=2\).

Step5: Check for extraneous solutions

• For \(x = 8\):

Left side of the original equation: \(x-3=8 - 3=5\)
Right side: \(\sqrt{4x-7}=\sqrt{4\times8 - 7}=\sqrt{32 - 7}=\sqrt{25}=5\)
Since \(5 = 5\), \(x = 8\) is a valid solution.

• For \(x = 2\):

Left side of the original equation: \(x - 3=2-3=-1\)
Right side: \(\sqrt{4x-7}=\sqrt{4\times2-7}=\sqrt{8 - 7}=\sqrt{1}=1\)
Since \(-1
eq1\), \(x = 2\) is an extraneous solution.

Answer:

8 (the option corresponding to 8, if the options are as given, the answer is the option with 8, e.g., if the option is "8" then the answer is 8)