(2x+3)^2-(5x-4)(5x+4)=(x+5)^2-(3x-1)(7x+2)-(x^2-x+1)

Jasonbradley

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Jun 16, 2024

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Solving the Equation: (2x+3)^2-(5x-4)(5x+4)=(x+5)^2-(3x-1)(7x+2)-(x^2-x+1)

This article will guide you through solving the equation: (2x+3)^2-(5x-4)(5x+4)=(x+5)^2-(3x-1)(7x+2)-(x^2-x+1). We will use algebraic manipulation to simplify the equation and ultimately find the value of 'x'.

Simplifying the Equation

Let's break down the equation step-by-step:

1. Expand the squares:

   • (2x+3)^2 = (2x+3)(2x+3) = 4x^2 + 12x + 9
   • (x+5)^2 = (x+5)(x+5) = x^2 + 10x + 25

2. Use the difference of squares pattern:

   • (5x-4)(5x+4) = (5x)^2 - (4)^2 = 25x^2 - 16

3. Expand the remaining products:

   • (3x-1)(7x+2) = 21x^2 + 5x - 2

4. Substitute the expanded terms back into the original equation:

   4x^2 + 12x + 9 - (25x^2 - 16) = x^2 + 10x + 25 - (21x^2 + 5x - 2) - (x^2 - x + 1)

5. Simplify by removing parentheses and combining like terms:

   4x^2 + 12x + 9 - 25x^2 + 16 = x^2 + 10x + 25 - 21x^2 - 5x + 2 - x^2 + x - 1 -21x^2 + 12x + 25 = -21x^2 + 6x + 26

6. Isolate 'x' on one side of the equation:

   -21x^2 + 12x + 25 + 21x^2 - 6x - 26 = 0 6x - 1 = 0

7. Solve for 'x':

   6x = 1 x = 1/6

Conclusion

Therefore, the solution to the equation (2x+3)^2-(5x-4)(5x+4)=(x+5)^2-(3x-1)(7x+2)-(x^2-x+1) is x = 1/6.