Animated Solution for Mathematics - Quadratic Equations: The value of 4+5+4+5+4+......∞1111
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Visualized Solution
Define the Variable
Let x=4+5+4+5+…111
Identify the Repeating Pattern
Notice that the pattern repeats after the first two terms: 4 and 5.
The part starting from the second 4 is identical to the original expression x.
Form the Recursive Equation
Substitute x into the repeating part: x=4+5+x11
Simplify the Inner Denominator
Focus on the term 5+x1.
By taking a common denominator, we get: 5+x1=x5x+1
Invert the Nested Fraction
Substitute the simplified denominator back: x=4+x5x+11
This simplifies to: x=4+5x+1x
Isolate the Fractional Term
Move 4 to the left side: x−4=5x+1x
Clear the Denominator
Multiply both sides by (5x+1):
(x−4)(5x+1)=x
Expand the Algebraic Expression
Expand the left side:
5x2+x−20x−4=x
Form the Standard Quadratic Equation
Combine like terms and move everything to one side:
5x2−19x−4=x
5x2−20x−4=0
Apply the Quadratic Formula
For ax2+bx+c=0, the roots are given by:
x=2a−b±b2−4ac
Substitute the Coefficients
Here, a=5, b=−20, c=−4.
Substitute these values:
x=2(5)−(−20)±(−20)2−4(5)(−4)
Calculate the Discriminant
Simplify the terms inside the square root:
x=1020±400+80
x=1020±480
Simplify the Radical Term
Factor out perfect squares from 480:
480=16×30=430
Final Simplification
Substitute the simplified radical back:
x=1020±430
x=2±5230
Conclusion and Key Takeaway
Since the original expression consists of positive terms, x>0.
Therefore, we reject the negative root.
Final Answer:x=2+5230
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The Sigma Insight: Solution of Quadratic Equations
The Infinite Mirror
A Journey into Continued Fractions
Imagine you are standing in a room with two parallel mirrors. You look into one, and you see an infinite corridor of reflections, each one identical to the last. This is exactly what we are dealing with in this problem.
We are looking at an infinite continued fraction:
x=4+5+4+5+…111
At first glance, it looks like a monster that will never end. But in mathematics, infinity is not a wall; it is a tool.
Phase 1
The Recursive Loop
The secret to taming this beast is to find the pattern. Look closely at the structure: we have a 4, then a 5, then another 4, then another 5. The pattern repeats perfectly.
If we define the entire expression as x, then the part of the fraction starting from the second 4 is also x. This is the magic of self-similarity.
We can rewrite our infinite expression as a simple recursive equation:
x=4+5+x11
Suddenly, the infinite has become finite. We have trapped the infinity inside a single variable.
Phase 2
The Algebraic Crucible
Now that we have our equation, it is time to do some heavy lifting. We need to solve for x.
Let's focus on the inner denominator: 5+x1. To combine these, we find a common denominator, which gives us x5x+1.
Now, substitute this back into our main equation:
x=4+x5x+11
Remember your rules of fractions: dividing by a fraction is the same as multiplying by its reciprocal. So, the expression becomes:
x=4+5x+1x
To make this easier to handle, let's move the 4 to the other side: x−4=5x+1x. Now, multiply both sides by (5x+1) to clear the denominator:
(x−4)(5x+1)=x
Phase 3
The Quadratic Resolution
Expand the left side carefully: 5x2+x−20x−4=x. Combine the like terms: 5x2−19x−4=x.
Finally, bring the x from the right side over to the left:
5x2−20x−4=0
We have arrived at a standard quadratic equation. To solve for x, we use the quadratic formula:
x=2a−b±b2−4ac
Here, a=5, b=−20, and c=−4. Plugging these in, we get:
x=2(5)20±(−20)2−4(5)(−4)
This simplifies to x=1020±400+80, which is:
x=1020±480
Phase 4
The Final Selection
We are almost there. We need to simplify 480. Since 480=16×30, we can pull out the 4: 480=430.
So, x=1020±430, which simplifies to:
x=2±5230
Now, we must make a choice. We have two possible values for x, but only one is physically meaningful. Since our original expression is a sum of positive terms, x must be positive.
The root 2−5230 is negative, so we reject it. Our final answer is:
x=2+5230
You have just conquered an infinite fraction by turning it into a simple quadratic. Keep this logic in your toolkit; whenever you see a repeating pattern, look for the self-similarity, and you will find the path to the solution.