sqrt(10-3x) + sqrt(x^2+5) = 7
$\displaystyle \sqrt{10-x}+\sqrt{x^2+5}=7$
The domain is $\displaystyle x \in (-\infty~,~10]$
The general procedure with such a type of equation is:
Square both sides - isolate remaining roots at one side - square again - isolate ... and so on:
$\displaystyle 10-x + 2\sqrt{10-x} \cdot \sqrt{x^2+5}+x^2+5 = 49~\implies~ 2\sqrt{-x^3+10x^2-5x+50}=-x^2+x+34$ ...... Square again:
$\displaystyle 4(-x^3+10x^2-5x+50)=x^4-2x^3-67x^2+68x+1156$ ...... Collect like terms:
$\displaystyle x^4+2x^3-107x^2+88x+956 = 0$
I don't know how to solve such an equation algebraically. So I used my computer which calculated 4 approximate solutions:
$\displaystyle \left| \begin{array}{l}x_1\approx -11.4101 \\ x_2 \approx -2.6234 \\ x_3 \approx 3.95199 \\ x_4 \approx 8.08147\end{array} \right.$
Since squaring is not an equivalent operation you must check these values by plugging them into the original equation and check if they satisfy the equation.
Better use a calculator
Hello,
You missed a 3...
$\displaystyle \sqrt{10-3x}+\sqrt{x^2+5}=7$
Domain is $\displaystyle \left]-\infty~,~\frac{10}{3}\right]$
Squaring :
$\displaystyle (10-3x)+(x^2+5)+2 \cdot \sqrt{(10-3x)(x^2+5)}=49$
$\displaystyle 2 \cdot \sqrt{(10-3x)(x^2+5)}=-x^2+3x+34$
Squaring :
$\displaystyle 4 (10-3x)(x^2+5)=x^4+9x^2+34^2-6x^3-68x^2+204x$
$\displaystyle 40x^2+200-12x^3-60x=x^4-6x^3-59x^2+204x+1156$
$\displaystyle 0=x^4+6x^3-99x^2+264x+956$
-2 is a solution
$\displaystyle 0=(x+2)(x^3+4x^2-107x+478)$
And it seems that the remaining has no real solution.