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Solve an equation with an extraneous solution

Solve x + 1 = 7 x + 15. . x + 1 = 7 x + 15. . ( x + 1) 2. = ( 7 x + 15) 2. . EXAMPLE 5. Solve an equation with an extraneous solution. Write original equation. Square each side. Expand left side and simplify right side. x 2 + 2 x + 1 = 7 x + 15.

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Solve an equation with an extraneous solution

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  1. Solve x + 1 = 7x + 15 .  x + 1 = 7x + 15  (x + 1)2 = ( 7x + 15)2  EXAMPLE 5 Solve an equation with an extraneous solution Write original equation. Square each side. Expand left side and simplify right side. x2 + 2x + 1 = 7x + 15 x2 – 5x – 14 = 0 Write in standard form. Factor. (x – 7)(x + 2) = 0 x – 7 = 0 or x + 2 = 0 Zero-product property Solve for x. x = 7 orx = –2

  2. Check x = 7 in the original equation. Check x = –2 in the original equation. x+ 1 = 7x+ 15 x+ 1 = 7x+ 15   7+ 1 –2+ 1 ?  = 7(–2) + 15 8 –1 = 7(7) + 15 ?  = 64 = 1 ? ?   8 = 8 –1 The only solution is 7. (The apparent solution 22 is extraneous.) ANSWER = 1 / EXAMPLE 5 Solve an equation with an extraneous solution CHECK 1

  3. Solve 2 2 METHOD 1 Solve using algebra. x + 2 + 1 = 3 – x   x + 2 + 1 x + 2 + 1   = 3 – x . = 3 – x   x + 2 +1 x + 2 +2   2 x + 2 EXAMPLE 6 Solve an equation with two radicals SOLUTION Write original equation. Square each side. = 3 – x Expand left side and simplify right side. = – 2x Isolate radical expression.

  4. x + 2 x + 2   2 EXAMPLE 6 Solve an equation with two radicals = –x Divide each side by 2. = ( –x)2 Square each side again. = x2 x + 2 Simplify. 0 = x2 – x – 2 Write in standard form. = (x – 2)(x + 1) 0 Factor. x – 2 = 0 x + 1 = 0 or Zero-product property. x = –1 Solve for x. x = 2 or

  5. x+ 2 +1 x+ 2 +1   2+ 2 +1  –1+ 2 +1  ? = 3 – (–1)  ? = 3 – x = 3 – 2 = 3 – x    1 +1  = 1 ? = 4 ?    4 +1 2 = 2 3 ANSWER The only solution is 1. (The apparent solution 2 is extraneous.) = –1 / EXAMPLE 6 Solve an equation with two radicals Check x = 2 in the original equation. Check x = – 1 in the original equation.

  6. Use: a graph to solve the equation. Use a graphingcalculator to graphy1 =+ 1andy2=. Thenfind the intersection points of the two graphs by using the intersect feature. You will find that the only point ofintersection is(21, 2).Therefore, 21 is the onlysolution of the equation x + 2 x + 2 + 1 = 3 – x 3 – x EXAMPLE 6 Solve an equation with two radicals METHOD 2

  7. x – 1 11.x – 1 x x = = 2 2 1 1 1 () 4 4 4 2 (x + )2 x = x2 –x + = x 5 x2 – x + = 0 4 1 1 1 1 4 4 2 4 for Examples 5 and 6 GUIDED PRACTICE Solve the equation. Check for extraneous solutions Write original equation. Square each side. Expand left side and simplify right side. 4x2– 5x + 1 = 0 Write in standard form. (4x – 1)(x – 1) = 0 Factor.

  8. x = Check x = 1 in the original equation. Check x = in the original equation. 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 x – x – = = x x 4 2 4 4 4 4 2 2 2 4 2 4 4 4 4 2 1 – x x 1 – = – – 1 16 = 1 1 1 4 4 2 ? ? ? ? = = = = for Examples 5 and 6 GUIDED PRACTICE or 4x – 1 = 0 x – 1 = 0 Zero-product property. x = 1 or Solve for x. The only solution is 1 (the apparent solution 1/4 is extraneous)

  9. 12. 10x + 9 = x + 3 10x + 9 = x + 3 ( ) 2 (x+ 3)2 10x + 9 ? = for Examples 5 and 6 GUIDED PRACTICE Solve the equation. Check for extraneous solutions Write original equation. Square each side. Expand right side and simplify left side. 10x + 9 = x2 + 6x +9 x2 – 4x = 0 Write in standard form. x (x – 4) = 0 Factor. or x = 0 (x – 4) = 0 Zero-product property. or x = 4 x = 0 Solve for x.

  10. Check x = 4 in the original equation. Check x = 0 in the original equation. 10x + 9 = x + 3 10x + 9 = x + 3 10x 0+ 9 0 + 3 10x 4+ 9 4 + 3 9 3 40 + 9 7 49 7 ? ? ? ? ? = = = = = for Examples 5 and 6 GUIDED PRACTICE 3 = 3 7 = 7 The solution are 4 and 0.

  11. 13. = = (x+ 7 ) ( ) 2 2 = 2x + 5 2x + 5 2x + 5 x + 7 x + 7 for Examples 5 and 6 GUIDED PRACTICE Solve the equation. Check for extraneous solutions Write original equation. Square each side. 2x + 5 = x+ 7 Simplify both the sides. x – 2 = 0 Simplify. x = 2 Simplify.

  12. = 2x + 5 x + 7 9 9 2.2 + 5 2 + 7 ? ? = = for Examples 5 and 6 GUIDED PRACTICE Check x = 2 in the original equation 3 = 3 The solution is 2.

  13. 14. = ( ) 2 = x + 6 –2 = x – 2 x – 2 x + 6 – 2 x – 2 ( ) x + 6 –2 2 x + 6 = 3 x + 6 – 4 x + 6 + 4 = x – 2 = – 12 – 4 x + 6 for Examples 5 and 6 GUIDED PRACTICE Solve the equation. Check for extraneous solutions Solve Write original equation. Square each side. Simplify each side.

  14. x + 6 = 3 x + 6 ( ) 2 = 32 for Examples 5 and 6 GUIDED PRACTICE Divide each side by –4. Square each side. x + 6 = 9 Simplify. x = 3 Simplify.

  15. x + 6 – 2 = x – 2 3 + 6 – 2 3 – 2 9 – 2 1 ? = ? = for Examples 5 and 6 GUIDED PRACTICE Check x =3 in the original equation 1 = 1

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