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January 7th, 2019, 02:31 AM   #1
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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
$\sqrt{x^2+3x+7}-\sqrt{x^2-3x+9}+2=0$
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 $50x^2-12x+130$ 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.
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January 7th, 2019, 04:20 AM   #2
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Quote:
Originally Posted by sigma View Post
$\sqrt{x^2+3x+7}-\sqrt{x^2-3x+9}+2=0$
Some work:

$\displaystyle a=x^2-3x+9,\quad b=x^2+3x+7$

$\displaystyle 2=\sqrt a-\sqrt b$

$\displaystyle 2\sqrt{ab}=(a+b)-4$

$\displaystyle 0=(a-b)^2-8(a+b)+16$

$\displaystyle 0=5x^2-6x-27$

I leave it to you to fill in and finish up.
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January 7th, 2019, 05:08 AM   #3
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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.
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January 7th, 2019, 05:20 AM   #4
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Quote:
Originally Posted by sigma View Post
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.

$x^2+m^2=2mx+(nx)^2$
n^2(x^2) - x^2 + 2mx - m^2 = 0

(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!
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January 7th, 2019, 05:49 AM   #5
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Quote:
Originally Posted by sigma View Post
Why is the equality equal to 2 in the beginning?
Look CAREFULLY at Greg's substitution to a and b:

sqrt(b) - sqrt(a) + 2 = 0

so: 2 = sqrt(a) - sqrt(b)

You're spending too much time on your iPhone
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January 7th, 2019, 08:26 AM   #6
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(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)
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January 7th, 2019, 09:23 AM   #7
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An alternative approach to greg1313's:
\begin{align}
\sqrt{x^2+3x+7} - \sqrt{x^2-3x+9} + 2 &= 0 \\
\sqrt{x^2-3x+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)} \\
2-6x &= 2\sqrt{x^2 + 3x + 7} + 2\sqrt{x^2 - 3x + 9} \\
1-3x &= \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 \\
(x-3)(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 $$a^2 - b^2 = (a-b)(a+b)$$
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.
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Last edited by v8archie; January 7th, 2019 at 09:42 AM.
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