Question

1. solve the following:

\sqrt{10 - 3x} = x

Explanation:

Step1: Square both sides

To eliminate the square root, we square both sides of the equation \(\sqrt{10 - 3x}=x\). Squaring the left side gives \(10 - 3x\) and squaring the right side gives \(x^{2}\), so we get the quadratic equation \(x^{2}+3x - 10 = 0\) (by rearranging \(10-3x=x^{2}\) to standard quadratic form \(x^{2}+3x - 10=0\)).

Step2: Factor the quadratic

We factor the quadratic equation \(x^{2}+3x - 10 = 0\). We need two numbers that multiply to \(- 10\) and add to \(3\). The numbers are \(5\) and \(-2\), so the equation factors as \((x + 5)(x - 2)=0\).

Step3: Solve for x

Setting each factor equal to zero gives \(x+5 = 0\) or \(x - 2=0\). Solving these gives \(x=-5\) or \(x = 2\).

Step4: Check for extraneous solutions

We check these solutions in the original equation. For \(x=-5\), the left side is \(\sqrt{10-3(-5)}=\sqrt{10 + 15}=\sqrt{25}=5\), and the right side is \(-5\). Since \(5
eq - 5\), \(x=-5\) is extraneous. For \(x = 2\), the left side is \(\sqrt{10-3(2)}=\sqrt{10 - 6}=\sqrt{4}=2\), and the right side is \(2\). So \(x = 2\) is a valid solution.

Answer:

\(x = 2\)