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Example:

Solve x2 + 5x + 6 = 0.This equation is already in the form "(quadratic) equals (zero)" but, unlike the previous example, this isn't yet factored. The quadratic must first be factored, because it is only when you MULTIPLY and get zero that you can say anything about the factors and solutions. You can't conclude anything about the individual terms of the unfactored quadratic (like the 5x or the 6), because you can add lots of stuff that totals zero.So the first thing I have to do is factor:x2 + 5x + 6 = (x + 2)(x + 3)Set this equal to zero:(x + 2)(x + 3) = 0Solve each factor: Copyright © Elizabeth Stapel 2002-2011 All Rights Reservedx + 2 = 0 or x + 3 = 0

x = –2 or x = – 3The solution to x2 + 5x + 6 = 0 is x = –3, –2Checking x = –3 and x = –2 in x2 + 5x + 6 = 0:[–3]2 + 5[–3] + 6 ?=? 0

9 – 15 + 6 ?=? 0

9 + 6 – 15 ?=? 0

15 – 15 ?=? 0

0 = 0[–2]2 + 5[–2] + 6 ?=? 0

4 – 10 + 6 ?=? 0

4 + 6 – 10 ?=? 0

10 – 10 ?=? 0

0 = 0So both solutions "check".

Solve x2 + 5x + 6 = 0.This equation is already in the form "(quadratic) equals (zero)" but, unlike the previous example, this isn't yet factored. The quadratic must first be factored, because it is only when you MULTIPLY and get zero that you can say anything about the factors and solutions. You can't conclude anything about the individual terms of the unfactored quadratic (like the 5x or the 6), because you can add lots of stuff that totals zero.So the first thing I have to do is factor:x2 + 5x + 6 = (x + 2)(x + 3)Set this equal to zero:(x + 2)(x + 3) = 0Solve each factor: Copyright © Elizabeth Stapel 2002-2011 All Rights Reservedx + 2 = 0 or x + 3 = 0

x = –2 or x = – 3The solution to x2 + 5x + 6 = 0 is x = –3, –2Checking x = –3 and x = –2 in x2 + 5x + 6 = 0:[–3]2 + 5[–3] + 6 ?=? 0

9 – 15 + 6 ?=? 0

9 + 6 – 15 ?=? 0

15 – 15 ?=? 0

0 = 0[–2]2 + 5[–2] + 6 ?=? 0

4 – 10 + 6 ?=? 0

4 + 6 – 10 ?=? 0

10 – 10 ?=? 0

0 = 0So both solutions "check".