68614 4 u 1 33 118614 4 now using the substitution

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Unformatted text preview: 3 = Þ 1 y -3 = 3 = 8 y Þ 1 y3 = = 1 1 Þ 1 y3 = 8 The two solutions to this equation are y = 1 and y = 1 Þ y = (1) 3 = 1 Þ æ 1 ö3 1 y=ç ÷ = 2 è8ø 1 1 . 2 [Return to Problems] (c) z - 9 z + 14 = 0 This one is a little trickier to see that it’s quadratic in form, yet it is. To see this recall that the exponent on the square root is one-half, then we can notice that the exponent on the first term is twice the exponent on the second term. So, this equation is in fact reducible to quadratic in form. Here is the substitution. u= z u2 = ( z) 2 =z The equation then becomes, u 2 - 9u + 14 = 0 ( u - 7 )( u - 2 ) = 0 u = 2, u = 7 Now go back to z’s. u=2: Þ u=7: z =2 Þ Þ z =7 Þ The two solutions for this equation are z = 4 and z = 49 z = ( 2) = 4 2 z = ( 7 ) = 49 2 [Return to Problems] (d) t 4 - 4 = 0 Now, this part is the exception to the rule that we’ve been using to identify equations that are reducible to quadratic in form. There is only one term with a t in it. However, notice that we can write the equation as, (t )...
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