Solving Radical Equations: Power Rule and Extraneous Solutions, Slides of Algebra

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§7.6 Radical

Equations

Review §

 Any QUESTIONS About

• §7.5 → Rational Exponents

 Any QUESTIONS About HomeWork

• §7.5 → HW-

MTH 55

Power Rule vs Radical Eqns

• Power Rule for Solving Radical Equations:

If BOTH SIDES of an equation are RAISED

TO THE SAME POWER, ALL solutions of

the original equation are ALSO solutions

of the NEW equation

Caveat PowerRule → Check

• CAUTION
• Read the power rule carefully; it does not

say that all solutions of the new equation

are solutions of the original equation. They

may or may not be…

• Solutions that do not satisfy the original

equation are called extraneous solutions;

they must be discarded.

 Thus the CHECK is CRITICAL

Example  Solve by PwrRule

• Solve Radical Equations: a) b) y = 12 3 x = − 4

 SOLUTION

a) b) ( ) 2 2 y = 12 y = 144 144 = 12 12 = 12 Check True ( ) (^ ) 3 3 3 x = − 4 x = − 64 Check 3 − 64 = − 4 − 4 = − 4 True

Example  Solve

 SOLUTION

4 x = x + 60 4 x = x + 60 ( ) ( ) 2 2 4 x = x + 60 ( ) 2 2 4 x = x + 60 16 x = x + 60 15 x = 60 x = 4

 Check

4 x = x + 60 4 4 = 4 + 60 4 2⋅ = 64 8 = 8

 4 Satisfies the

original Eqn,

so 4 is verified

as a Solution

Example  Solve

 SOLUTION

x − 5 = x + 7 x − 5 = x + 7 ( ) (^) ( ) 2 2 x − 5 = x + 7 2 x − 10 x + 25 = x + 7 2 x − 11 x + 25 = 7 2 x − 11 x + 18 = 0 ( x − 2)( x − 9) = 0 x − 2 = 0 or x − 9 = 0 x = 2 x = 9 Square both sides. Use FOIL or Formula. Subtract x from both sides. Factor. Use the zero-products theorem. Subtract 7 from both sides. The TENTATIVE Solutions

Example  Solve

x − 5 = x + 7

 Check BOTH Tentative Solutions

x = 2 2 − 5 = 2 + 7 − 3 = 9 − 3 ≠ 3 False. x = 9 9 − 5 = 9 + 7 4 = 16 4 = 4 True.

 Because 2 does not check, it is an

extraneous solution. The only soln is 9

Example  Solve

 SOLUTION  Check

 So 13 checks. The solution set is {13}

 4 x + 3 + 3 = 5. 4 x + 3 + 3 = 5 4 x + 3 = 2 ( ) 4 4 4 x + 3 = 2 x + 3 = 16 x = 13 4 x + 3 + 3 = 5 ( ) 4 13 + 3 + 3 = 5 4 16 + 3 = 5 2 + 3 = 5 5 = 5

Example  Solve

 SOLN

 Check

 So 9 checks. The

solution set is {9}

 m + 3 = 9 m = 6 ( ) 2 2 m = 6 m = 36 Using the Power Rule Isolate the variable radical m + 3 = 9

m + 3 = 9 6 + 3 = 9

Example  Solve

 x = x + 5 + 1

 Check BOTH Tentative Solutions

x = x + 5 + 1

4 + 5 + 1 9 + 1 3+ 4 4

x = x + 5 + 1

− + 1 5 + 1 4 + 1 2+ − 1 − (^1) 

 In this Case 4 checks while −1 does

NOT. The solution set is {4}

Example  Solve

 SOLUTION

3 x + 4 − 2 = 0 3 3 x + 4 = 2 ( ) 3 3 2 3 3 x + 4 = 3 x + 4 = 8 3 x = 4 x = 4 / 3 3 3 x + 4 − 2 = 0

 CHECK

3 4 2 0 3 x + − = 4 2 0 3 4 3 ? (^3) + − =      4 4 2 0 ? 3

• − = 8 2 0 ? 3 − = 2 − 2 = (^0) 

All Done for Today

Life Expectancy

Bruce Mayer, PE Licensed Electrical & Mechanical Engineer [email protected] Chabot Mathematics

Appendix

r − s ≡^ (^ r − s )( r^ + s )

2 2