# Solve Real Number $x$ That Satisfies Equation

• MHB
• anemone
In summary, solving for a real number $x$ that satisfies an equation means finding the value of $x$ that makes the equation true. This involves isolating the variable $x$ by using algebraic operations until its value is determined. There can be more than one real number $x$ that satisfies an equation, and if there is no real number $x$ that satisfies an equation, then the equation has no solutions. This concept is applicable in various real-life situations and is an essential skill in fields such as science, engineering, and finance.
anemone
Gold Member
MHB
POTW Director
Find a real number $x$ that satisfy the following equation:

$x = \sqrt{(x-667)(x-736)}+\sqrt{(x-736)(x-928)}+\sqrt{(x-928)(x-667)}$

Optimistic guess: suppose that $x-667$, $x-736$ and $x-928$ are all perfect squares, say $x-667 = a^2$, $x-736 = b^2$ and $x-928 = c^2$. Then $a^2 - b^2 = 736 - 667 = 69$. But if $a$ and $b$ are integers with $(a+b)(a-b) = 69$ then either $a+b=23$ and $a-b = 3$ or $a+b=69$ and $a-b = 1$. In the first of those cases, $a=13$ and $b=10$. Then $x-667 = 13^2 = 169$, so that $x= 836$. But if $x = 836$ then $c^2 = 836 - 928$, which is negative. So that solution does not work.

I then tried the other possible case, with $a+b=69$ and $a-b = 1$, but that did not lead to a solution either. So it looked as though there was not going to be a solution in integers.

But then the fact that $c^2$ turned out to be negative in my first attempt at a solution made me think about the possibility that $x-667$, $x-736$ and $x-928$ might all be negatives of squares. In that case, my first solution would have to be adjusted, to make $x-667 = -100$ so that $x=567$. Then $x-736 = -169 = -13^2$ and $x-928 = -361 = -19^2$.

Finally, if $x = 567$ then \begin{aligned}\sqrt{(x-667)(x-736)}+\sqrt{(x-736)(x-928)}+\sqrt{(x-928)(x-667)} &= \sqrt{(-100)(-169)} + \sqrt{(-169)(-361)} + \sqrt{(-361)(-100)} \\ &= 10*13 + 13*19 + 19*10 \\ &= 130+247+190 \\ &= 567.\end{aligned}

If I see correctly, there should be 2 distinct solutions over real numbers for the equation, one of which around $$x \approx 1184.13$$. Does the finding of this include to the challenge too?

Hi Theia, as long as you find one real $x$ value that satisfies the given equation, then that $x$ will be an answer to the problem.

We can write the original equation in the form
$x = \sqrt{a} + \sqrt{b} + \sqrt{c},$
where
\begin{align*} a &= (x - 667)(x - 736) \\ b &= (x - 736)(x - 928) \\ c &= (x - 928)(x - 667). \end{align*}
By solving the $\sqrt{c}$ from the equation and squaring one obtains
$c = x^2 - 2x\sqrt{b} - 2x\sqrt{a} + b + 2\sqrt{a}\sqrt{b} + a.$
Let's now move all terms with $\sqrt{b}$ dependence to left hand side and others to right hand side:
$2\sqrt{b} \left( \sqrt{a} - x \right) = -x^2 + 2x\sqrt{a} + c - b - a.$
After squaring only $\sqrt{a}$ is left. Collecting those terms and squaring one can write \begin{align*} & x^4 - 4x^3\sqrt{a} - 2cx^2 - 2bx^2 + 6ax^2 + 4cx\sqrt{a} + 4bx\sqrt{a} - 4ax\sqrt{a} \\ & \quad + c^2 - 2bc - 2ac + b^2 - 2ab + a^2 = 0 \qquad \Leftrightarrow \end{align*}
\begin{align*} 4x\sqrt{a} \left( x^2 - c - b + a \right) = &\ x^4 - 2cx^2 - 2bx^2 + 6ax^2 + c^2 - 2bc \\ & - 2ac + b^2 - 2ab + a^2, \end{align*}
and finally
$\begin{multline} x^8 - 4\left( a + b + c \right) x^6 + \left( 6c^2 + \left( 4b + 4a \right) c + 6b^2 + 4ab + 6a^2 \right ) x^4 \\ + 4\left( \left( a + b - c \right) c^2 + \left( b^2 - 10ab + a^2 \right) c - b^3 + ab^2 + a^2b - a^3\right) x^2 \\ + c^4 - 4\left( a + b \right) c^3 + \left( 6b^2 + 4ab + 6a^2 \right) c^2 + \left( -4b^3 + 4ab^2 + 4a^2b \right. \\ \left. - 4a^3 \right) c + b^4 - 4ab^3 + 6a^2b^2 - 4a^3b + a^4 = 0. \end{multline}$
Let's now substitute expressions for $a$, $b$ and $c$ and simplify:
\begin{align*} & 29156245504x^5 - 84494799470592x^4 + 90809255028129792x^3 \\ & - 46242678657209335808x^2 + 11296854647375577219072x \\ & - 1067552764176992047202304 = 0. \end{align*}
This quintic can be reduced because we know its one solution, namely $x = 567$ by Opalg. Hence we obtain a quartic:
$x^4-2331x^3+1792896x^2-569457920x+64576528128 = 0.$
The solution of this quartic is a boring algebraic procedure. First one needs to substitute $x = t + 2331/4$ to get rid of the $x^3$ term. Hence the quartic equation can be written as
$t^4 - \frac{1957515}{8}t^2 - \frac{504331747}{8}t - \frac{1124218036179}{256} = 0.$
Next one needs to move $t^1$ and $t^0$ terms to right hand side, and manipulate the left hand side to a square:
$\left( t^2 - \frac{1957515}{16} \right) ^2 = \frac{504331747}{8}t + \frac{1239020752851}{64}.$
To get the right hand side to a square too, one can add an unknown number $q$ to the equation such that left hand side stays a square. In other words: $\left( t^2 - \frac{1957515}{16} + q \right)^2 = 2qt^2 + \frac{504331747}{8}t + q^2 - \frac{1957515q}{8} + \frac{1239020752851}{64}.$
Left hand side is a quadratic function of $t$ now. For this to be a perfect square, one must write its discriminant (with respect to $t$) equal to $0$.
$512q^3 - 125280960q^2 + 9912166022808q - 254350511032072009 = 0.$
To solve this cubic one first substitutes $q = p + 652505/8$ to remove $q^2$ term. That gives the reduced cubic
$p^3 - 597930816p - 2945222125792 = 0.$
Now, as the cubic has 3 real roots (check out yourself!), one can manipulate the reduced cubic using the formula for cosine, $4\cos ^3 \theta - 3\cos \theta = \cos 3\theta$. To do this, one substitutes $p = u\cos \theta$ and chooses $u$ accordingly:
\begin{align} 4\cos ^3 \theta - \frac{2391723264}{u^2}\cos \theta &= \frac{11780888503168}{u^3} \quad \Rightarrow \\ \frac{2391723264}{u^2} &= 3 \quad \Leftrightarrow \\ u &= 10672\sqrt{7} \quad \Rightarrow \\ \cos 3\theta &= \frac{206879}{149408\sqrt{7}} \end{align}
Now one can solve $\theta$ and hence finally $q$ by going backwards the substitutions that were made:
\begin{align*} \theta &= \frac{1}{3}\arccos \left( \frac{206879}{149408\sqrt{7}} \right) + \frac{2n\pi}{3} \quad \Rightarrow \\ p &= 10672\sqrt{7} \cos \left( \frac{1}{3}\arccos \left( \frac{206879}{149408\sqrt{7}} \right) + \frac{2n\pi}{3} \right) \quad \Rightarrow \\ q &= \frac{652505}{8} + 10672\sqrt{7} \cos \left( \frac{1}{3}\arccos \left( \frac{206879}{149408\sqrt{7}} \right) + \frac{2n\pi}{3} \right) . \end{align*}
Hence the approximate solutions for the parameter $q$ are:
\begin{align*} q_{n=0} &= q_0 \approx 108182.1818025683 \\ q_{n=1} &= q_1 \approx 60098.83251836192 \\ q_{n=2} &= q_2 \approx 76408.36067906974. \end{align*}
Now, as the number $q$ is known, one can continue the solving of the quartic equation and write its left hand side to a square:
$\left( t^2 + q - \frac{1957515}{16} \right) ^2 = 2q\left( t + \frac{504331747}{32q} \right) ^2.$
After moving terms to right hand side, one can use the formula $a^2 - b^2 = (a+b)(a-b)$:
$\begin{multline} \left[ \frac{32q^2 - 3915030q + 504331747\sqrt{2q}}{32q} + \sqrt{2q} \ t + t^2 \right] \\ \cdot \left[ \frac{32q^2 - 3915030q - 504331747\sqrt{2q}}{32q} - \sqrt{2q} \ t + t^2 \right] = 0. \end{multline}$
Now these 2 quadratic equations are trivial to solve and one can write down the solutions for the original quartic (and hence the original square root equation) by going backwards the substitutions: \begin{align} x_1 &= \frac{2331}{4} - \frac{ 4\sqrt{2q^3} + \sqrt{2}\sqrt{ -16q^3 + 3915030q^2 - 504331747\sqrt{2q^3}}}{8q} \\ x_2 &= \frac{2331}{4} - \frac{ 4\sqrt{2q^3} - \sqrt{2}\sqrt{ -16q^3 + 3915030q^2 - 504331747\sqrt{2q^3}}}{8q} \\ x_3 &= \frac{2331}{4} + \frac{ 4\sqrt{2q^3} - \sqrt{2}\sqrt{-16q^3 + 3915030q^2 + 504331747\sqrt{2q^3}}}{8q} \\ x_4 &= \frac{2331}{4} + \frac{ 4\sqrt{2q^3} + \sqrt{2}\sqrt{-16q^3 + 3915030q^2 + 504331747\sqrt{2q^3}}}{8q} , \end{align}
where $q = q_1$ (or $q_0$ or $q_2$, but preferred the smallest absolute value). A closer look to the solutions $x_i$ tells, only $x_4 \approx 1184.131\ldots$ is correct and rest are false solutions due to the squaring of the original equation.

P.S. Posted this from phone... Line breaks can be placed poorly. Sorry about that! (Whew)

## What is the definition of a real number?

A real number is any number that can be found on the number line, including both positive and negative numbers, fractions, and decimals. It is a continuous set of numbers with no gaps or jumps.

## What does it mean to solve an equation?

To solve an equation means to find the value of the variable that makes the equation true. This is done by performing operations on both sides of the equation until the variable is isolated on one side and the solution is on the other side.

## What is the difference between an equation and an expression?

An equation is a mathematical statement that shows the relationship between two or more quantities, while an expression is a combination of numbers, variables, and operations that does not have an equal sign. Equations can be solved to find the value of the variable, while expressions cannot.

## How do I know if a real number satisfies an equation?

To determine if a real number satisfies an equation, you can substitute the value into the equation and see if it makes the equation true. If it does, then the number is a valid solution to the equation.

## What is the process for solving an equation with multiple steps?

The process for solving an equation with multiple steps is to first simplify both sides of the equation by combining like terms and using the distributive property. Then, isolate the variable on one side by performing inverse operations, such as addition and subtraction, until the variable is alone. Finally, check your solution by substituting it back into the original equation.

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