# Solve the following ?

## $\sqrt{3 {x}^{2} - 7 x - 30} + \sqrt{2 {x}^{2} - 7 x - 5} = x + 5$

Jun 25, 2018

$x = 6$

#### Explanation:

Given:

$\sqrt{3 {x}^{2} - 7 x - 30} + \sqrt{2 {x}^{2} - 7 x - 5} = x + 5$

We can get this into polynomial form as follows:

Square both sides (noting that this may introduce extraneous solutions) to get:

$\left(3 {x}^{2} - 7 x - 30\right) + 2 \sqrt{\left(3 {x}^{2} - 7 x - 30\right) \left(2 {x}^{2} - 7 x - 5\right)} + \left(2 {x}^{2} - 7 x - 5\right) = {x}^{2} + 10 x + 25$

That is:

$5 {x}^{2} - 14 x - 35 + 2 \sqrt{6 {x}^{4} - 35 {x}^{3} - 26 {x}^{2} + 245 x + 150} = {x}^{2} + 10 x + 25$

Subtracting ${x}^{2} + 10 x + 25 + 2 \sqrt{6 {x}^{4} - 35 {x}^{3} - 26 {x}^{2} + 245 x + 150}$ from both sides, this becomes:

$4 {x}^{2} - 24 x - 60 = - 2 \sqrt{6 {x}^{4} - 35 {x}^{3} - 26 {x}^{2} + 245 x + 150}$

Dividing both sides by $2$, this becomes:

$2 {x}^{2} - 12 x - 30 = - \sqrt{6 {x}^{4} - 35 {x}^{3} - 26 {x}^{2} + 245 x + 150}$

Squaring both sides (noting that this may introduce extraneous solutions) this becomes:

$4 {x}^{4} - 48 {x}^{3} + 24 {x}^{2} + 720 x + 900 = 6 {x}^{4} - 35 {x}^{3} - 26 {x}^{2} + 245 x + 150$

Subtracting the left hand side from the right, this becomes:

$2 {x}^{4} + 13 {x}^{3} - 50 {x}^{2} - 475 x - 750 = 0$

By the rational roots theorem, any rational zeros of this quartic are expressible in the form $\frac{p}{q}$ for integers $p , q$ with $p$ a divisor of the constant term $- 750$ and $q$ a divisor of the coefficient $2$ of the leading term.

Hence the only possible rational zeros are:

$\pm \frac{1}{2} , \pm 1 , \pm \frac{3}{2} , \pm 2 , \pm \frac{5}{2} , \pm 3 , \pm 5 , \pm 6 , \ldots , \pm 750$

Trying each in turn, we find $x = - \frac{5}{2}$ is a root and hence $\left(2 x + 5\right)$ a factor:

$0 = 2 {x}^{4} + 13 {x}^{3} - 50 {x}^{2} - 475 x - 750$

$\textcolor{w h i t e}{0} = \left(2 x + 5\right) \left({x}^{3} + 4 {x}^{2} - 35 x - 150\right)$

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 $x = - 5$ is another zero and $\left(x + 5\right)$ a factor:

${x}^{3} + 4 {x}^{2} - 35 x - 150 = \left(x + 5\right) \left({x}^{2} - x - 30\right) = \left(x + 5\right) \left(x - 6\right) \left(x + 5\right)$

So the zeros of the quartic are:

$- 5 , - 5 , - \frac{5}{2} , 6$

Trying each of these in the original equation, we find the only valid solution is $x = 6$

Jun 25, 2018

$x = 6$

#### Explanation:

Given:

$\sqrt{3 {x}^{2} - 7 x - 30} + \sqrt{2 {x}^{2} - 7 x - 5} = x + 5$

$\implies \sqrt{3 {x}^{2} - 7 x - 30} - 5 = x - \sqrt{2 {x}^{2} - 7 x - 5}$

Squaring both sides

=>3x^2-7x-30+25-2*5sqrt(3x^2-7x-30) =x^2+2x^2-7x-5 -2xsqrt(2x^2-7x-5)

=>3x^2-7x-5-2*5sqrt(3x^2-7x-30) =3x^2-7x-5 -2xsqrt(2x^2-7x-5)

=>(cancel(3x^2-7x-5))-2*5sqrt(3x^2-7x-30) =(cancel(3x^2-7x-5) )-2xsqrt(2x^2-7x-5)

=>5sqrt(3x^2-7x-30) =xsqrt(2x^2-7x-5)

Again squaring

=>25(3x^2-7x-30) =x^2(2x^2-7x-5)

$\implies 2 {x}^{2} - 7 {x}^{3} - 80 {x}^{2} + 175 x + 750 = 0$

By trial we get two zeros for $x = \pm 5$ . So ${x}^{2} - 25$ is a factor of LHS.

Now

$\implies 2 {x}^{2} - 7 {x}^{3} - 80 {x}^{2} + 175 x + 750 = 0$

$\implies 2 {x}^{4} - 50 {x}^{2} - 7 {x}^{3} + 175 x - 30 {x}^{2} + 750 = 0$

$\implies 2 {x}^{2} \left({x}^{2} - 25\right) - 7 x \left({x}^{2} - 25\right) - 30 \left({x}^{2} - 25\right) = 0$

$\implies \left({x}^{2} - 25\right) \left(2 {x}^{2} - 7 x - 30\right) = 0$

$\implies \left({x}^{2} - 25\right) \left(2 {x}^{2} - 12 x + 5 x - 30\right) = 0$

$\implies \left({x}^{2} - 25\right) \left(2 x \left(x - 6\right) + 5 \left(x - 6\right)\right) = 0$

$\implies \left({x}^{2} - 25\right) \left(2 x + 5\right) \left(x - 6\right) = 0$

So four zeros are

for $x = + 5 , - 5 , - \frac{5}{2} \mathmr{and} 6$

Among these only $x = 6$ satisfies the given equation.

Hence $x = 6$