Squared Plus Squared Is Squared

This solution looks into some mathematical approaches - geometric and algebraic.

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Geometric Approach

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Recall that the Pythagorean Theorem is $a^2 + b^2 = c^2$ where $a$ and $b$ are side lengths of the triangle, and $c$ is the hypotenuse.

The given equation can be expressed as $(2x - 1)^2 + \left(2(x + 3)\right)^2 = (2x + 7)^2$

Looking at each of the squared terms individually, we can see that the equation resembles the Pythagorean Theorem. In this case, we can relate this to the triangle of side lengths $(2x - 1)$ , $2(x + 3)$ and $(2x + 7)$ , where $(2x + 7)$ is the longest side of the triangle aka the hypotenuse. Here is the following illustration:

Note: For the solution, we allow the right triangle in the Cartesian plane, so we don't neglect triangles with negative side lengths. Otherwise, the solutions are not complete.

Let $y = 2x - 1$ , so the new equation is $y^2 + (y + 7)^2 = (y + 8)^2$

Visually, $5$ - $12$ - $13$ satisfies the given property, where $y = 5$ . Considering negative numbers of $y$ , $y = -3$ (which gives $3$ - $4$ - $5$ ) must also work!

Thus, $\boxed{x = -1, 3}$ . By noticing the side length differences, the method turns out to be very nice and elegant. :)

It is also possible, but not necessary (and possibly not rigorous!), to transform this problem into the system of equations. Set $a = y$ , $b = y + 7$ and $c = y + 8$ , so $\begin{array}{rl} -a + b &= 7\\ -a + c &= 8\\ -b + c &= 1 \end{array}$ where the triplets are $3-4-5$ and $5-12-13$ .

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Algebraic Approach

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From the previous approach, we simplified into smaller coefficients in the equation, which can easily be solved algebraically. The resulting equation after combining like terms is $y^2 - 2y - 15 = 0$

We can also solve for $x$ without having to perform the variable substitution. These methods are all great to solve the problem. :)

$(2x-1)^2+4(x+3)^2=(2x+7)^2$

Expand and combine like terms.

$4x^2-4x+1+4(x^2+6x+9)=4x^2+28x+49$

$-4x+1+4x^2+24x+36=28x+49$

$4x^2+24x-4x-28x+1+36-49=0$

$4x^2-8x-12=0$

$x^2-2x-3=0$

By factoring, we have

$(x-3)(x+1)=0$

$x=3$

$x=-1$

Check by substituting:

when $x=3$

$(2x-1)^2+4(x+3)^2=(2x+7)^2$

$[2(3)-1]^2+4(3+3)^2=[2(3)+7]^2$

$25+144=169$

$169=169$ (true)

when $x=-1$

$(2x-1)^2+4(x+3)^2=(2x+7)^2$

$[2(-1)-1]^2+4(-1+3)^2=[2(-1)+7]^2$

$9+16=25$

$25=25$ (true)

$\therefore$ The solutions are $\boxed{3}$ and $\boxed{-1}$ .