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# Sulalgtrig7e Isg 1 4

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### Transcript

• 1. Objective 1.4.1 Radical Equations
• 2. To solve an equation with a radical 2x −1 − 3 = 4 First isolate the radical +3 +3 This means to get any terms not under the square root on the other 2 2 side of the equal sign Now square both sides 2x −1 = 7 You must square the whole side NOT each term. A square 2 x − 1 = 49 Now solve for x "undoes" or cancels a square root +1 +1 You MUST check this answer 2 x = 50 2( 25) − 1 − 3 = 4 Since you squared both sides of the equation, negatives disappear. It is x = 25 4=4 possible to get anIt checks! doesn't work answer that when you plug it back in
• 3. Let's try another one: 1 ( 2 x + 1) 3 +1 = 0 -1 First isolate the radical Remember that the 1/3 -1 Now since it is same thing power means the a 1/3 as a cube root. power this means the 1 3 3 ( 2 x + 1) 3= −1 same as a cube root so cube both sides 2 x + 1 = −1 Now solve for x -1 -1 Let's check this answer 2 x = −2 32( − 1) + 1 + 1 = 0 x = −1 0 = 0 It checks!
• 4. One more to see extraneous solution: 3 x + 1 =that − 3find algebraically but DOES isolated a solution x you The radical is already NOT make a true statement when you substitute it back 2 into the equation. 2 Square both sides 3x + 1 = x − 3 You must square the whole side NOT each term. This must be FOILed 3 x + 1 = x − 6 x + 9 You MUST check 2 Since you have a quadratic these answersan x x −getxeverything on one side = 0 and see if 2 9 +8 = 0 equation (has 2 3 (8) + 1 = 1 − 3 term) ( x − 8)( x − 1) = 0 you can factor this 3 1 +1 = 8−3 x = 8, x = 1 It Doesn't work! 2=− checks! 5 = 5 2 Extraneous
• 5. Objective 1.4.2 Equations in Quadratic Form The "u" Substitution Method
• 6. x − 5x + 4 = 0 4 2 Before we solve the above equation, let's solve a quadratic equation that we know how to solve. u − 5u + 4 = 0 2 Factor ( u − 4)( u − 1) = 0 Set each factor = 0 and solve u = 4, u = 1 Let's use this to solve the original equation by letting u = x2.
• 7. x − 5x + 4 = 0 4 2 If u = x2 then square both sides and get u2 = x4. Substitute u and u2 for x2 and x4. u − 5u + 4 = 0 2 Factor ( u − 4)( u − 1) = 0 Set each factor = 0 and solve u = 4, u = 1 x = 4, x = 1 2 2 Now that we've solved for u we have to re-substitute to get x back. Remember u = x2 so let's substitute. Solve for x by square-rooting both sides and don't forget the ± x = ±2, x = ±1
• 8. You can determine if an equation is of quadratic form where you can use the "u" substitution method if you call the middle variable and power u and then square it and get the first term's variable and power. 1 1 (z ) = z 4 1 1 2 2 z 2 − 4z 4 +4=0 u − 4u + 4 = 0 2 So let u = z1/4 and get u2 = z1/2. Substitute u and u2 for z1/4 and z1/2. ( u − 2)( u − 2) = 0 u=2 Factor & set each factor = 0 and solve 1 Solve for z by raising both u=z =2 4 1 sides to the 4th power ( z ) = ( 2) 4 4 4 z = 16
• 9. Let's try one more. Call the middle variable u and then square it to see if you get the first term's variable. (x ) = x 3 2 6 x − 7x − 8 = 0 6 3 So let u = x3 and get u2 = x6. u − 7u − 8 = 0 2 Substitute u and u2 for x3 and x6. ( u − 8)( u + 1) = 0 Factor & set each factor = 0 and solve u = 8, u = −1 x = 8, x = −1 3 3 Solve for x by taking the cube root of both sides x = 2, x = −1