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on Mar 12, 2010

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• Quadratics come in this form: ax 2 + bx + c = 0 Sometimes it is factorable and finding the value(s) of x is easy. What do you do when it isn't factorable? Can you still solve for x?
• 2 + bx + c = 0 ax 2 + bx + c = 0 ax a a a a 2 x + bx +c=0 a a
• x 2 + bx +c=0 a a x 2 + bx =0-c a a 2 2= 2 x + bx + a ( ) b 2a -c + a ( ) b 2a
• 2 2 ( x+ b 2a ) = -c + a ( ) b 2a 2 2 ( x+ b 2a ) = -c + a ( ) b 2a 2 ( ) x+ b =± -c + b 2a a 2a
• 2 ( ) x+ b =± -c + b 2a a 2a 2 x =± -c + a ( ) b 2a - b 2a
• 2 x= ± -c + a ( )b 2a - b 2a 2 x= - ( ) b ± -c + b 2a a 2a 2 x= - ( ) b ± 2a -c + b 2a 2a a 2a
• 2 x= - ( ) b ± 2a -c + b 2a 2a a 2a 2 ( ) x= - b ± 2a -c + b a 2a 2a
• 2 ( ) x= - b ± 2a -c + b a 2a 2a 2 2 2 x= - b ± - c (2a) + a ( ) b 2a (2a) 2a
• 2 2 2 x= - b ± - c (2a) + a ( ) b 2a (2a) 2a 2 2 2 x= - b ± - c 4a + a ( ) b 4a 2 4a 2a
• 2 2 2 x= - b ± - c 4a + a ( ) b 4a 2 4a 2a 2 2 2 x= - b ± - c 4a + a ( ) b 4a 2 4a 2a
• 2 2 2 x= - b ± - c 4a + a ( ) b 4a 2 4a 2a 2 x= - b ± - c 4a + b 2a
• 2 x = -b ± b - 4ac 2a
• 2 7x + 14x - 3 = 0 What are the roots of the above function 2 x = -b ± b - 4ac 2a 2 7x + 14x - 3 = 0
• 2 7x + 14x - 3 = 0 2 x = -14 ± 14 - 4(7)(-3) 2(7) x = -14 ± 196 + 84 14
• x = -14 ± 280 14 x = -14 ± 16.73 14
• x = -14 ± 16.73 14 x1 = -14 + 16.73 x2 = -14 - 16.73 14 14 x1 = 2.73 x2 = -30.73 14 14 x1 = .2 x2 = -2.2