Solve the following ?
#sqrt(3x^2-7x-30)+sqrt(2x^2-7x-5)=x+5#
2 Answers
Explanation:
Given:
#sqrt(3x^2-7x-30)+sqrt(2x^2-7x-5) = x+5#
We can get this into polynomial form as follows:
Square both sides (noting that this may introduce extraneous solutions) to get:
#(3x^2-7x-30)+2sqrt((3x^2-7x-30)(2x^2-7x-5))+(2x^2-7x-5) = x^2+10x+25#
That is:
#5x^2-14x-35+2sqrt(6x^4-35x^3-26x^2+245x+150) = x^2+10x+25#
Subtracting
#4x^2-24x-60 = -2sqrt(6x^4-35x^3-26x^2+245x+150)#
Dividing both sides by
#2x^2-12x-30=-sqrt(6x^4-35x^3-26x^2+245x+150)#
Squaring both sides (noting that this may introduce extraneous solutions) this becomes:
#4x^4-48x^3+24x^2+720x+900 = 6x^4-35x^3-26x^2+245x+150#
Subtracting the left hand side from the right, this becomes:
#2x^4+13x^3-50x^2-475x-750=0#
By the rational roots theorem, any rational zeros of this quartic are expressible in the form
Hence the only possible rational zeros are:
#+-1/2, +-1, +-3/2, +-2, +-5/2, +-3, +-5, +-6, ..., +-750#
Trying each in turn, we find
#0 = 2x^4+13x^3-50x^2-475x-750#
#color(white)(0) = (2x+5)(x^3+4x^2-35x-150)#
Note that the remaining cubic does not factor by grouping, since the ratio between the first and second terms is different from the ratio between the third and fourth terms.
Using the rational roots theorem again and continuing, to try roots, we find that
#x^3+4x^2-35x-150 = (x+5)(x^2-x-30) = (x+5)(x-6)(x+5)#
So the zeros of the quartic are:
#-5, -5, -5/2, 6#
Trying each of these in the original equation, we find the only valid solution is
Explanation:
Given:
#sqrt(3x^2-7x-30)+sqrt(2x^2-7x-5) = x+5#
Squaring both sides
Again squaring
By trial we get two zeros for
Now
So four zeros are
for
Among these only
Hence