Question

1. x - 3 = √(15 - x) solve for extraneous

Explanation:

Step1: Square both sides to eliminate square root

To solve the equation \( x - 3=\sqrt{15 - x} \), we first square both sides of the equation. Squaring the left side \((x - 3)^2\) and the right side \((\sqrt{15 - x})^2\) gives us:

$$ (x - 3)^2=15 - x $$

Step2: Expand the left - hand side

Using the formula \((a - b)^2=a^{2}-2ab + b^{2}\) where \(a = x\) and \(b = 3\), we expand \((x - 3)^2\):

$$ x^{2}-6x + 9=15 - x $$

Step3: Rearrange into standard quadratic form

Move all terms to the left - hand side of the equation to get a quadratic equation in standard form \(ax^{2}+bx + c = 0\). Add \(x\) to both sides and subtract \(15\) from both sides:

$$ x^{2}-6x+x + 9 - 15=0 $$

Simplify the equation:

$$ x^{2}-5x - 6=0 $$

Step4: Factor the quadratic equation

We factor the quadratic equation \(x^{2}-5x - 6\). We need two numbers that multiply to \(-6\) and add up to \(-5\). The numbers are \(-6\) and \(1\). So,

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

Step5: Solve for x

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

• If \(x - 6=0\), then \(x = 6\).
• If \(x + 1=0\), then \(x=-1\).

Step6: Check for extraneous solutions

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

• For \(x = 6\):

Left - hand side: \(x-3=6 - 3=3\)
Right - hand side: \(\sqrt{15 - x}=\sqrt{15 - 6}=\sqrt{9}=3\)
Since \(3 = 3\), \(x = 6\) is a valid solution.

• For \(x=-1\):

Left - hand side: \(x - 3=-1-3=-4\)
Right - hand side: \(\sqrt{15 - x}=\sqrt{15-(-1)}=\sqrt{16}=4\)
Since \(-4
eq4\), \(x=-1\) is an extraneous solution.

Answer:

The solution to the equation \(x - 3=\sqrt{15 - x}\) is \(x = 6\) (after checking for extraneous solutions, \(x=-1\) is rejected).