Question

solve \\(sqrt{5x - 4} = x - 2\\)
hint: after squaring, check both solutions in the original equation.
skill sheet: solving radical equations + \check for extraneous solutions,\ p 3–4

a (x = 1) only

b (x = 8) only

c (x = 1) and (x = 8)

d no solution

Explanation:

Step1: Square both sides to eliminate radical

To solve the equation \(\sqrt{5x - 4}=x - 2\), we first square both sides. Squaring the left side gives us \(5x - 4\) (since \((\sqrt{a})^2=a\) for \(a\geq0\)), and squaring the right side \((x - 2)^2\) expands to \(x^{2}-4x + 4\) using the formula \((a - b)^2=a^{2}-2ab + b^{2}\) where \(a = x\) and \(b = 2\). So we get the equation:
\[5x-4=x^{2}-4x + 4\]

Step2: Rearrange into quadratic equation

We move all terms to one side to form a quadratic equation. Subtract \(5x\) and add \(4\) to both sides:
\[x^{2}-4x + 4-5x + 4 = 0\]
Simplify the like terms:
\[x^{2}-9x + 8=0\]

Step3: Factor the quadratic equation

We factor the quadratic equation \(x^{2}-9x + 8 = 0\). We need two numbers that multiply to \(8\) and add up to \(- 9\). The numbers are \(-1\) and \(-8\). So the factored form is:
\[(x - 1)(x - 8)=0\]

Step4: Solve for x

Using the zero - product property, if \((x - 1)(x - 8)=0\), then either \(x-1 = 0\) or \(x - 8=0\). Solving these gives \(x = 1\) or \(x = 8\).

Step5: Check for extraneous solutions

• For \(x = 1\): Substitute into the original equation \(\sqrt{5x-4}=x - 2\). The left - hand side is \(\sqrt{5(1)-4}=\sqrt{1}=1\), and the right - hand side is \(1 - 2=-1\). Since \(1

eq - 1\), \(x = 1\) is an extraneous solution.

• For \(x = 8\): Substitute into the original equation. The left - hand side is \(\sqrt{5(8)-4}=\sqrt{40 - 4}=\sqrt{36}=6\), and the right - hand side is \(8 - 2 = 6\). Since \(6=6\), \(x = 8\) is a valid solution.

Answer:

B. \(x = 8\) only