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One solution of #x^3+(2-i)x^2+(-4-3i)x+(1+i)=0# is #x=1+i#. Find the only positive real solution for #x#?

Hint: #1-i# is not necessarily a root because the coefficients are imaginary

Answer 1

Lily Anderson

#-3/2+sqrt(13)/2#

#x^3+(2-i)x^2+(-4-3i)x+(1+i)#

#=x^3-(1+i)x^2+3x^2+(-4-3i)x+(1+i)#

#=x^3-(1+i)x^2+3x^2-3(1+i)x-x+(1+i)#

#=x^2(x-(1+i))+3x(x-(1+i))-1(x-(1+i))#

#=(x^2+3x-1)(x-(1+i))#

So the other two roots are the roots of #x^2+3x-1 = 0#, which we can find using the quadratic formula:

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

#color(white)(x) = -3/2+-sqrt(13)/2#

So the positive real root is:

#-3/2+sqrt(13)/2#

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.