How do you solve #-\frac { 1 } { 2 } ( t - 3 ) + t ^ { 2 } = 0#?

1 Answer
Oct 14, 2016

#t=1/4(1+-sqrt(23)color(white)(.)i)#

Explanation:

Given:#" "-1/2(t-3)+t^2=0#

Note that when multiplying or dividing, when the signs are different, you end up with a minus. When the signs are the same you end up with a plus.
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#color(blue)("Dealing with the brackets first")#
Multiply everything inside the brackets by #(-1/2)#

#color(brown)((-1/2)xx(+t)" = "-1/2t) larr" "#minus times plus gives minus

#color(brown)((-1/2)xx(-3)" = "+3/2) larr" "#minus times minus gives plus

#color(brown)("Putting it all together:")#

#color(blue)(-1/2t+3/2+t^2=0)#
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Write this as:

#t^2-1/2t+3/2=0#

compare to standardised form:

#ax^2+bx+c=0" "->" "x=(-b+-sqrt(b^2-4ac))/(2a)#

where in this case #x->t; a=1"; "b=-1/2"; "c=3/2#

#=> t=(1/2+-sqrt((-1/2)^2-4(1)(3/2)))/(2(1))#

#t=1/4+-(sqrt(1/4-6))/2#

#t=1/4+-sqrt(-23/4)/sqrt(4)#

#t=1/4+-sqrt(-23/4xx1/4)#

#t=1/4+-1/4sqrt(23)sqrt(-1)#

but #sqrt(-1) = i#

#t=1/4(1+-sqrt(23)color(white)(.)i)#