Solve for x : (x+1)^2 = 2x^2 - 5x + 11
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We have to solve (x+1)^2=2x^2-5x+11 for x.
(x+1)^2=2x^2-5x+11
open the brackets
=> x^2 + 2x + 1 = 2x^2 - 5x + 11
=> x^2 - 7x + 10 = 0
=> x^2 - 5x - 2x + 10 = 0
=> x(x - 5) - 2(x - 5) = 0
=> (x - 2)(x - 5) = 0
x - 2 = 0 => x = 2
and x - 5 = 0 => x = 5
The values of x are x = 5 and x = 2
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The first step is to expand the square from the left side, using the formula:
(a+b)^2 = a^2 + 2ab + b^2
(x+1)^2 = x^2 + 2x + 1
We'll re-write the given equation:
x^2 + 2x + 1 = 2x^2 - 5x + 11
We'll move all terms to the right side:
0 = 2x^2 - x^2 - 5x - 2x + 11 - 1
We'll combine like terms and we'll use symmetrical property:
x^2 - 7x + 10 = 0
We'll apply quadratic formula:
x1 = [7 + sqrt(49 - 40)]/2
x1 = (7+sqrt9)/2
x1 = (7+3)/2
x1 = 5
x2 = (7-3)/2
x2 = 2
The quadratic equation has the following solutions: x1 = 5 and x2 = 2.
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