# Solve for x : (x+1)^2 = 2x^2 - 5x + 11

justaguide | College Teacher | (Level 2) Distinguished Educator

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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

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

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

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.