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Solve Simultaneous Equations Enter ValuesSolving Quadratic EquationsLooking at an example: You are given the question x2+5x+4=0 In the above example, a=1 (If there is no number before the x then we can assume that thenumber is 1) b=5 c=4 What we then have to do is find out what x could be equal to in order to satisfy the equation and therefore make it true. In this example, x can either equal -4 or -1 . We don't know yet how we came to this answer, but let's show that both of these answers do work. Putting -4 into the above equation: x2+5x+4=0 (-4)2 + (5*-4) + 4 = 0 (-4)*(-4) + (5*-4) + 4 = 0 16 + (-20) + 4 = 0 16-20+4=0 -4+4=0 0=0 This shows that -4 can be a solution for x Putting -1 into the above equation: x2+5x+4=0 (-1)2 + (5*-1) + 4 = 0 (-1)*(-1) + (5*-1) + 4 = 0 1 + (-5) + 4 = 0 1-5+4=0 -4+4=0 0=0 This shows that -1 can also be a solution for x Therefore, there are two possible solutions, -1 and -4. They must both begiven as an answer to obtain full marks. Once you have obtained the possible solutions for x, is is always necessary to checkthem in the above way |
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