help me solve this square root equation

**cityismine** · May 17th 2008, 07:50 PM

sqrt(10-3x) + sqrt(x^2+5) = 7

**earboth** · May 17th 2008, 08:25 PM

Originally Posted by cityismine

sqrt(10-3x) + sqrt(x^2+5) = 7

$\sqrt{10-x}+\sqrt{x^2+5}=7$

The domain is $x \in (-\infty~,~10]$

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:

$10-x + 2\sqrt{10-x} \cdot \sqrt{x^2+5}+x^2+5 = 49~\implies~ 2\sqrt{-x^3+10x^2-5x+50}=-x^2+x+34$ ...... Square again:

$4(-x^3+10x^2-5x+50)=x^4-2x^3-67x^2+68x+1156$ ...... Collect like terms:

$x^4+2x^3-107x^2+88x+956 = 0$

I don't know how to solve such an equation algebraically. So I used my computer which calculated 4 approximate solutions:
$\left| \begin{array}{l}x_1\approx -11.4101 \\ x_2 \approx -2.6234 \\ x_3 \approx 3.95199 \\ x_4 \approx 8.08147\end{array} \right.$

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

**Moo** · May 18th 2008, 02:09 AM

Hello,

Originally Posted by earboth

$\sqrt{10-{\color{red}3}x}+\sqrt{x^2+5}=7$

[...]

You missed a 3...

$\sqrt{10-3x}+\sqrt{x^2+5}=7$

Domain is $\left]-\infty~,~\frac{10}{3}\right]$

Squaring :

$(10-3x)+(x^2+5)+2 \cdot \sqrt{(10-3x)(x^2+5)}=49$

$2 \cdot \sqrt{(10-3x)(x^2+5)}=-x^2+3x+34$

Squaring :

$4 (10-3x)(x^2+5)=x^4+9x^2+34^2-6x^3-68x^2+204x$

$40x^2+200-12x^3-60x=x^4-6x^3-59x^2+204x+1156$

$0=x^4+6x^3-99x^2+264x+956$

-2 is a solution

$0=(x+2)(x^3+4x^2-107x+478)$

And it seems that the remaining has no real solution.

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