How do you solve #-3x^2 - 5 = 22#?

2 Answers
Jun 3, 2016

Answer:

How do you solve #-3x^2-5=22#?

Explanation:

Since there is no "linear" term, that is a term with an "x" variable that is not squared, I would isolate the "#x^2#" term like so:
#-3x^2-5+5=22+5# <add 5 to both sides>
#-3x^2=27#
#(-3x^2)/(-3)=(27/-3)# <divide both sides by -3>
#x^2=-9# which can be solved with imaginary numbers!

Take the square root of both sides: #sqrt(x^2)=sqrt(-9)#
And don't forget the #+-# sign: #x = +- 3i#
That is, there are no real solutions!

Graphically, this means that the function graph of f(x) = #-3x^2-5# will not intersect the graph of f(x) = 22 on the real number plane.

Jun 3, 2016

Answer:

I have taken you to a point where you can take over.

Explanation:

#color(blue)("Comment")#
It is a matter of looking at whole number factors of the #x^2# term and the constant. Then you play around with them until you find the combination that works. Sometimes the factors that work are not whole numbers in which case you will most likely need to use the formula to find them.
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#color(blue)("Answering the question")#

Add #3x^2" and "5# to both sides giving:

#3x^2+5x+22=0#

Whole number factors of 22 can only be 2 and 11
Whole number factors of 3 can only be 3 and 1

Starting point. We know we have: #(3x+?)(x+?)#

Considering combinations for the constant of 11

Try this one: #(3x+2)(x+11) = 3x^2+11x+2x+22color(red)(" Fail")#
Try this one: #(3x+11)(x+2) = 3x^2+6x+11x+22 color(red)(" Fail")#

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

#color(brown)("Ok! looks as though we need to use the formula")#

standard form #y=ax^2+bx+c# where #a=3"; "b=5"; "c=22#

and #x=(-b+-sqrt(b^2-4ac))/(2a)#

#=>x=(-5+-sqrt((-5)^2-4(3)(22)))/(2(3))#

#=>x=(-5+-sqrt(-239))/(6)#

As soon as you see that there is a square root of a negative number then the curve does not cross the x-axis. Consequently there is no solution where #x in RR# (Real numbers)

Thus the only solutions are for #x in CC# (Complex numbers)

#=>x=(-5+-[sqrt(239)color(white)(.)xxi])/(6)#

I will let you finish this off.
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Tony B