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January 7th, 2019, 03:31 AM  #1 
Newbie Joined: Oct 2018 From: arizona Posts: 6 Thanks: 0  Equations with Roots and Exponents
I'm struggling with the equation below. I've tried putting it into quadratics form to use the quadratics formula, but it has been too computationally arduous for me. I'm not sure where else to start. $x^2+m^2=2mx+(nx)^2$ Since I'm posting making this post, I may as well ask for help on another problem, which is $\sqrt{x^2+3x+7}\sqrt{x^23x+9}+2=0$ I've tried isolating one of the roots on the left side and squaring both sides, but that has not worked. I've tried squaring both sides initially, and so far that has not worked (I end up with $50x^212x+130$ doing this, somehow). I've tried representing the roots as terms with fractional exponents, and no luck there either. I must be misusing arithmetic properties or something. Note: for both equations, the goal is to find all real solutions. Complex solutions are to be excluded. 
January 7th, 2019, 05:20 AM  #2 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,900 Thanks: 1094 Math Focus: Elementary mathematics and beyond  Some work: $\displaystyle a=x^23x+9,\quad b=x^2+3x+7$ $\displaystyle 2=\sqrt a\sqrt b$ $\displaystyle 2\sqrt{ab}=(a+b)4$ $\displaystyle 0=(ab)^28(a+b)+16$ $\displaystyle 0=5x^26x27$ I leave it to you to fill in and finish up. 
January 7th, 2019, 06:08 AM  #3 
Newbie Joined: Oct 2018 From: arizona Posts: 6 Thanks: 0 
Why is the equality equal to 2 in the beginning? Shouldn't it be 2 since we would have to subtract it to get it on the other side? I don't understand the third line of reasoning that is \displaystyle 2\sqrt{ab}=(a+b)4 However, I did work with the polynomial in the end and got the same answer my textbook got, so I know you're right. I just don't understand why you're right. 
January 7th, 2019, 06:20 AM  #4  
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,803 Thanks: 970  Quote:
(n^2  1)x^2 + 2mx  m^2 = 0 So: a = n^2  1, b = 2m, c = m^2 Finish it.....and happy new year!  
January 7th, 2019, 06:49 AM  #5 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,803 Thanks: 970  
January 7th, 2019, 09:26 AM  #6 
Global Moderator Joined: Dec 2006 Posts: 20,105 Thanks: 1907 
(nx)²  (x  m)² = 0 $\implies$ (nx  x + m)(nx + x  m) = 0 (n²  1)x² + 2mx  m² ≡ ((n  1)x + m)((n + 1)x  m) 
January 7th, 2019, 10:23 AM  #7 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,561 Thanks: 2562 Math Focus: Mainly analysis and algebra 
An alternative approach to greg1313's: \begin{align} \sqrt{x^2+3x+7}  \sqrt{x^23x+9} + 2 &= 0 \\ \sqrt{x^23x+9}  \sqrt{x^2+3x+7} &= 2 &\text{(Eqn 1)} \\ (x^2  3x + 9)  (x^2 + 3x + 7) &= 2\sqrt{x^2 + 3x + 7} + 2\sqrt{x^2  3x + 9} &\text{(Eqn 2)} \\ 26x &= 2\sqrt{x^2 + 3x + 7} + 2\sqrt{x^2  3x + 9} \\ 13x &= \sqrt{x^2 + 3x + 7} + \sqrt{x^2  3x + 9} &\text{(Eqn 3)} \\ \text{(Eqn 1)} + \text{(Eqn 3)} \implies 2\sqrt{x^2  3x + 9} &= 3  3x \\ 4(x^2  3x + 9) &= 9  18x + 9x^2 \\ 5x^2  6x  27 &= 0 \\ \tfrac15\big((5x)^2  6(5x)  135\big) &= 0 \\ \tfrac15\big((5x)15\big)\big((5x) + 9\big) &= 0 \\ (x3)(5x+9) &= 0 \end{align} For $\text{(Eqn 2)}$ we multiplied by $\big( \sqrt{x^2 + 3x + 7} + \sqrt{x^2  3x + 9} \big)$ using the identity $$a^2  b^2 = (ab)(a+b)$$ This has the advantage that we only square the once. Each time we square the equation there is a risk that we add solutions that may not satisfy the original equation, so we have to validate every solution that we obtain: $$\sqrt{3^2+3\times 3 + 7}\sqrt{3^2  3\times 3 + 9} = \sqrt{25}  \sqrt{9} = 5  3 = 2$$ So $x=3$ is not a solution. Last edited by v8archie; January 7th, 2019 at 10:42 AM. 

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equations, exponents, quadratics, roots 
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