Radical Equations and Extraneous Roots Answers

Study the steps shown to solve the given equation.30 – 2x = x2 – 6x + 9 0 = x2 – 4x – 21 0 = (x + 3)(x – 7)Based on the above work, possible solutions of the equation are –7 and –3–7 and 3–3 and 73 and 7.

x =✔ 8.250.751.75X 2.25no real solutions

Solve:To solve this equation, first add 2 tomultiply by 3 on X square ✔ cube both sides.

Study the steps shown to solve the given equation.30 – 2x = x2 – 6x + 9 0 = x2 – 4x – 21 0 = (x + 3)(x – 7)Based on the above work, possible solutions of the equation are –7 and –3–7 and 3✔ –3 and 73 and 7.

Check the answers in the original equation. The extraneous solution is x =.

Complete the work shown to find a possible solution of the equation. A possible solution of the equation is ⇒ 14.

The value 14 is an extraneous solution, so the equation has no real solutions.Complete the sentence below.You can tell that the equation has no real solutions from the second line, because taking the square root results in the ⇒ principal, positive root only.