Determine the value of variable: (x+5)^2-4(4x+5)=0
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We have to solve (x+5)^2-4(4x+5)=0
(x+5)^2 - 4(4x+5) = 0
open the brackets
x^2 + 10x + 25 - 16x - 20 = 0
=> x^2 - 6x + 5 = 0
=> x^2 - 5x - x + 5 = 0
=> x( x - 5) - 1(x - 5) = 0
=> (x - )(x - 5) = 0
Therefore x = 1 and x = 5
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To solve the equation, we'll expand the square and we'll remove the brackets.
We'll expand the square using the formula:
(a+b)^2 = a^2 + 2ab + b^2
Now, we'll expand (x+5)^2:
(x+5)^2 = x^2 + 10x + 25
The equation will become:
x^2 + 10x + 25 - 16x - 20 = 0
We'll combine like terms:
x^2 - 6x + 5 = 0
We'll apply quadratic formula;
x1 = [6+sqrt(36 - 20)]/2
x1 = (6+4)/2
x1 = 5
x2 = (6-4)/2
x2 = 1
The solutions of the equation are {1 ; 5}.
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