Question

question 40 of 41
what is the solution to the equation below?
\\(\sqrt{x + 3} = x - 3\\)

a. 5
b. 3
c. 6
d. 4

Explanation:

Step1: Square both sides to eliminate the square root

To solve the equation \(\sqrt{x + 3}=x - 3\), we first square both sides of the equation. Squaring the left side \((\sqrt{x + 3})^2\) gives \(x + 3\), and squaring the right side \((x - 3)^2\) using the formula \((a - b)^2=a^2-2ab + b^2\) (where \(a = x\) and \(b = 3\)) gives \(x^2-6x + 9\). So we get the new equation:

$$x + 3=x^2-6x + 9$$

Step2: Rearrange the equation to form a quadratic equation

Subtract \(x\) and \(3\) from both sides of the equation \(x + 3=x^2-6x + 9\) to set the equation to zero:

$$x^2-6x + 9-x - 3=0$$

Simplify the left - hand side:

$$x^2-7x + 6=0$$

Step3: Factor the quadratic equation

We factor the quadratic equation \(x^2-7x + 6 = 0\). We need to find two numbers that multiply to \(6\) and add up to \(-7\). The numbers are \(-1\) and \(-6\). So the factored form is:

$$(x - 1)(x - 6)=0$$

Step4: Solve for x

Using the zero - product property, if \((x - 1)(x - 6)=0\), then either \(x-1 = 0\) or \(x - 6=0\).

• If \(x-1 = 0\), then \(x = 1\). But we need to check this solution in the original equation. Substitute \(x = 1\) into \(\sqrt{x + 3}=x - 3\):

Left - hand side: \(\sqrt{1 + 3}=\sqrt{4}=2\)
Right - hand side: \(1-3=-2\)
Since \(2
eq - 2\), \(x = 1\) is an extraneous solution.

• If \(x - 6=0\), then \(x = 6\). Check this solution in the original equation:

Left - hand side: \(\sqrt{6 + 3}=\sqrt{9}=3\)
Right - hand side: \(6-3 = 3\)
Since \(3 = 3\), \(x = 6\) is a valid solution.

Answer:

C. 6