(2t)(t+2)=1
I get 2 different answers from quadratic form. and factoring.
I get 2 different answers from quadratic form. and factoring.
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In order to solve by factoring or the quadratic formula, the equation has to be equal to 0. If you leave it equal to 1, your answer will not be correct.
(2t)(t + 2) = 1 (distribute)
2t^2 + 4t = 1 (subtract 1 from both sides)
2t^2 + 4t - 1 = 0
At this point, you could solve by factoring or by using the quadratic formula. However, 2t^2 + 4t - 1 does not factor, so you have to solve using the quadratic formula. In this case, a = 2, b = 4, and c = -1.
t = [-b ± √(b^2 - 4ac)]/(2a)
t = [-4 ± √(4^2 - 4(2)(-1))]/(2*2)
t = [-4 ± √(16 + 8)]/4
t = (-4 ± √24)/4
t = (-4 ± 2√6)/4
t = (-2 ± √6)/2 <===ANSWER
(2t)(t + 2) = 1 (distribute)
2t^2 + 4t = 1 (subtract 1 from both sides)
2t^2 + 4t - 1 = 0
At this point, you could solve by factoring or by using the quadratic formula. However, 2t^2 + 4t - 1 does not factor, so you have to solve using the quadratic formula. In this case, a = 2, b = 4, and c = -1.
t = [-b ± √(b^2 - 4ac)]/(2a)
t = [-4 ± √(4^2 - 4(2)(-1))]/(2*2)
t = [-4 ± √(16 + 8)]/4
t = (-4 ± √24)/4
t = (-4 ± 2√6)/4
t = (-2 ± √6)/2 <===ANSWER
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If your answers from factorising were t=0 or t= -2, you are forgetting that the equation must equal zero to use factorisation.
So expand brackets, subtract 1 (from both sides) and attempt to factorise. You will quickly realise that this quadratic doesn't factorise and so the formula (or completing the square) are the only ways to solve it.
Solutions are t = 1/2 ( (+-) sqrt(6)-2), or t= 0.224745, t= -2.224745
So expand brackets, subtract 1 (from both sides) and attempt to factorise. You will quickly realise that this quadratic doesn't factorise and so the formula (or completing the square) are the only ways to solve it.
Solutions are t = 1/2 ( (+-) sqrt(6)-2), or t= 0.224745, t= -2.224745