#x^3+x-10=0#? In detail, please

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Answer 1 · 2018-01-30T22:43:21

#x=2# This is the correct answer if the question is looking for only real number answers.

Let's use a approach that only requires basic knowledge.

Let's manipulate this equation a bit.

#x^3+x-10=0#

#x^3+x=10#

#x(x^2+1)=10#

Let's assume(since problems like this one usually are) that #x# is an integer.

More specifically, a positive one, since there is no way for both #x# and #x^2+1# to be negative.

#(1,10)# #(2,5)# are the only factors of #10#

When #x=1#, #x^2+1# equals #2#, not #10#. Therefore, #x!=1#

When #x=2#, #x^2+1# is indeed #5.# Therefore, #x=2#

Answer 2 · 2018-01-30T22:53:42

Read below.

We try to find the #x# intercept of #x^3+x-10#

After trial and error, we see that #(2,0)# is the #x#-intercept.

We now divide #x^3+x-10# by #x-2# to get #x^2+2x+5#

We now try to find the #x# intercept of #x^2+2x+5#.

Let's use the quadratic formula on this.

#(-b+-sqrt(b^2-4(a)(c)))/(2(a))#

=>#(-2+-sqrt(2^2-4(1)(5)))/(2(1))#

=>#(-2+-sqrt(4-20))/2#

=>#(-2+-sqrt(-16))/2#

=>#(-2+-4i)/2#

=>#-1+-2i#

We now see that: #x^3+x-10=(x-2)(x+1-2i)(x+1+2i)#

Therefore, #(x-2)(x+1-2i)(x+1+2i)=0#

The answers are: #2,(-1+2i),(-1-2i)#