Find all the real solutions of the equation.

square root of the quantity (x+7), minus the square root of the quantity (x+2)

equals the square root of the quantity (x-1), minus the square root of the

quantity (x-2).

Thanks for your help.

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- Jan 31st 2006, 07:04 PMmkfadeaway11Need PreCalculus Help
Find all the real solutions of the equation.

square root of the quantity (x+7), minus the square root of the quantity (x+2)

equals the square root of the quantity (x-1), minus the square root of the

quantity (x-2).

Thanks for your help. - Jan 31st 2006, 08:59 PMearbothQuote:

Originally Posted by**mkfadeaway11**

if you want to eliminate the square roots, you have to square both sides of your equation. Make sure you use the binomial formula correctly and be aware that the transformation of the equation does**not**give an equivalent equation:

$\displaystyle \sqrt{x+7} - \sqrt{x+2}=\sqrt{x-1}-\sqrt{x-2}$

LHS of equation:

$\displaystyle x+7-2 \cdot \sqrt{x+7} \cdot \sqrt{x+2} + x+2 $

RHS of equation:

$\displaystyle x-1-2 \cdot \sqrt{x-1} \cdot \sqrt{x-2}+x-2$

$\displaystyle 12=2 \sqrt{(x+7)(x+2)}-2\sqrt{(x-1)(x-2)}$

Square both sides again:

LHS of equation: 144

RHS of equation:

$\displaystyle 4(x+7)(x+2)-8 $$\displaystyle \sqrt{x^4+6x^3-11x^2-24x+28}+4(x-1)(x-2)$

Isolate the square root $\displaystyle 8\sqrt{x^4+6x^3-11x^2-24x+28}$on one side of your equation:

$\displaystyle 8 \cdot \sqrt{...}=8x^2+24x+64$

Divide first by 8 and then square both sides of your equation. Use $\displaystyle (a+b+c)^2=a^2+b^2+c^2+2ab+2ac+2bc$ with your RHS!

After a whole bunch of transformation you'll get:

$\displaystyle 36x^2+72x+36=0$

$\displaystyle 36 \cdot (x+1)^2=0$

To solve this equation is certainly no problem for you.

You have to prove whether your solution fits into the original equation or not, because you have made often no equavelant transformation!

Bye - Jan 31st 2006, 11:46 PMCaptainBlackQuote:

Originally Posted by**earboth**

original equation imaginary and the LHS real.

RonL - Feb 1st 2006, 02:20 AMCaptainBlackQuote:

Originally Posted by**mkfadeaway11**

$\displaystyle \sqrt{x+7}-\sqrt{x+2}=\sqrt{x-1}-\sqrt{x-2}$.

Squaring both sides and simplifying:

$\displaystyle 6-\sqrt{x-7} \sqrt(x+2)=-\sqrt{x-1} \sqrt{x-2}$.

Squaring again and simplifying:

$\displaystyle \sqrt{x+7}\sqrt{x+2}=4+x$.

Squaring again:

$\displaystyle (x+7)(x+2)=(4+x)^2$,

expanding and simplifying give:

$\displaystyle x=2$.

Substituting this back into the original equation confirms that this is

indeed a solution, and as we have not lost any solutions in our manipulations

we have that it is the only solution.

RonL - Feb 1st 2006, 07:26 AMearbothQuote:

Originally Posted by**CaptainBlack**

you're right. I've forgotten the 144 on the LHS of the equation. I'm awfully sorry!

This time: red hot ears.

Bye