# 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