(a-5)^2+(b-c)^2+(c-d)^2+(b+c+d-9)^2=0

Suzhoumeite

4 min read

·

Jul 03, 2024

37

Share

Solving the Equation (a-5)^2+(b-c)^2+(c-d)^2+(b+c+d-9)^2=0

In this article, we will explore the solution to the equation:

$(a-5)^2+(b-c)^2+(c-d)^2+(b+c+d-9)^2=0$

This equation may seem complex at first, but by using the properties of squares and some algebraic manipulations, we can simplify it and find the solution.

Step 1: Expand the Squares

Let's start by expanding each square:

$(a-5)^2=a^2-10a+25$ $(b-c)^2=b^2-2bc+c^2$ $(c-d)^2=c^2-2cd+d^2$ $(b+c+d-9)^2=b^2+2bc+2bd-18b+c^2+2cd-18c+d^2-18d+81$

Now, let's plug these expansions back into the original equation:

$a^2-10a+25+b^2-2bc+c^2+c^2-2cd+d^2+b^2+2bc+2bd-18b+c^2+2cd-18c+d^2-18d+81=0$

Step 2: Combine Like Terms

Next, let's combine like terms:

$a^2-10a+25+2b^2-2bc+2c^2-2cd+2d^2-18b-18c-18d+81=0$

Step 3: Simplify the Equation

Now, let's simplify the equation by combining the constants:

$a^2-10a+2b^2-2bc+2c^2-2cd+2d^2-18b-18c-18d+106=0$

Step 4: Factor the Equation

Unfortunately, this equation cannot be factored easily. However, we can try to find a solution by setting each term equal to zero and solving for each variable.

Case 1: a-5=0

$a-5=0\Rightarrow a=5$

Case 2: b-c=0

$b-c=0\Rightarrow b=c$

Case 3: c-d=0

$c-d=0\Rightarrow c=d$

Case 4: b+c+d-9=0

$b+c+d-9=0\Rightarrow b+c+d=9$

Conclusion

The equation $(a-5)^2+(b-c)^2+(c-d)^2+(b+c+d-9)^2=0$ can be simplified and solved by setting each term equal to zero and solving for each variable. The solutions are:

• $a=5$
• $b=c$
• $c=d$
• $b+c+d=9$

These solutions satisfy the original equation. Note that there may be other solutions as well, but these are the ones we can find using this method.