Using the Pythagorean theorem how do you find the unknown lengths A=x, B=2x-3, C=2x+3?

Hint: This question is from the topic of algebra. In this question, we have to find the lengths of A, B, and C. In solving this question, we will first understand about Pythagorean Theorem. After that, we will take the values A, B, and C and substitute them in the formula of Pythagorean Theorem. After solving the further question, we will get the value of x. After that, we will get the value of sides.

Complete step by step solution:
Let us solve this question.
In this question, we have to find the value of A, B, and C using Pythagorean Theorem, where A=x, B=2x-3, and C=2x+3.
So, let us first understand the Pythagorean Theorem by the following:

Here, we can see that hypotenuse side is the longest side that is (2x+3), perpendicular side is (x), and base side is (2x-3).
Now, in the formula of Pythagorean Theorem that is
\[{{\left( \text{hypotenuse side} \right)}^{2}}\text{ = }{{\left( \text{perpendicular side} \right)}^{2}}+{{\left( \text{base side} \right)}^{2}}\]
We will substitute the value of A, B, and C.
Hence, we can write
\[\Rightarrow {{\left( 2x+3 \right)}^{2}}\text{ = }{{\left( x \right)}^{2}}+{{\left( 2x-3 \right)}^{2}}\]
Using the formula \[{{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}\], we can write the above equation as
\[\Rightarrow {{\left( 2x \right)}^{2}}+2\times 2x\times 3+{{\left( 3 \right)}^{2}}\text{ = }{{x}^{2}}+{{\left( 2x-3 \right)}^{2}}\]
Now, using the formula \[{{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}\], we can write the above equation as
\[\Rightarrow {{\left( 2x \right)}^{2}}+2\times 2x\times 3+{{\left( 3 \right)}^{2}}\text{ = }{{x}^{2}}+{{\left( 2x \right)}^{2}}-2\times 2x\times 3+{{\left( 3 \right)}^{2}}\]
The above equation can also be written as
\[\Rightarrow 4{{x}^{2}}+12x+9\text{ = }{{x}^{2}}+4{{x}^{2}}-12x+9\]
\[\Rightarrow 4{{x}^{2}}+12x+9-\left( 4{{x}^{2}}-12x+9 \right)={{x}^{2}}\]
The above equation can also be written as
\[\Rightarrow 12x+12x={{x}^{2}}\]
\[\Rightarrow 24x={{x}^{2}}\]
The above equation can also be written as
\[\Rightarrow {{x}^{2}}=24x\]
\[\Rightarrow {{x}^{2}}-24x=0\]
Taking x common in the left side of the equation, we will get
\[\Rightarrow x\left( x-24 \right)=0\]
Here, we can say that x=0 or x-24=0
Or, we can say that
x=0 or x=24.
As one of the sides is x, so the value of x cannot be written as zero. Hence, we will take the value of x as 24.
Therefore,
\[A=x=24\]
\[B=2x-3=2\times 24-3=48-3=45\]
\[C=2x+3=2\times 24+3=48+3=51\]
We have found the unknown lengths using Pythagorean Theorem. The unknown lengths are:
A=24, B=45, and C=51

Note: As this question is from the topic of algebra, so we should have a better knowledge in the same topic. We should know about the Pythagorean Theorem. Let us understand the same from the following:

The formula is: \[{{\left( \text{hypotenuse side} \right)}^{2}}\text{ = }{{\left( \text{perpendicular side} \right)}^{2}}+{{\left( \text{base side} \right)}^{2}}\]
And, also remember the following formulas to solve this type of question easily:
\[{{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}\]
\[{{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}\]