Question

(x + 1)^2 = 3x – 1 solving Quadratics ?

Answer 1

Step-by-step explanation:

$(x + 1)^{2} = 3x - 1 \\ \\ \therefore \: {x }^{2} + 2x + 1 = 3x - 1 \\ \\ \therefore \: {x }^{2} + 2x + 1 - 3x + 1 = 0\\ \\ \therefore \: {x }^{2} + 2x - 3x + 1 - 1 = 0 \\ \\ \therefore \: {x }^{2} - x = 0 \\ \\ \therefore \: x(x - 1) = 0 \\ \\ \therefore \: x = 0 \: \: or \: \: x - 1 = 0 \\ \\ \therefore \: x = 0 \: \: or \: \: x = 1 \\ \\ \huge{ \red{ \boxed{\therefore \: x = \{0, \: \: 1 \}}}}$

Answer 2

Answer: x = - 1 or x = 2

Step-by-step explanation:

The given quadratic equation is expressed as

(x + 1)^2 = 3x – 1

Expanding the brackets, it becomes

(x + 1)(x + 1) = 3x - 1

x² + x + x + 1 = 3x - 1

x² + 2x - 3x + 1 + 1 = 0

x² - x + 2 = 0

We would find two numbers such that their sum or difference is -x and their product is 2x^2. The two numbers are x and 2x. Therefore,

x² + x - 2x + 2 = 0

x(x + 1) - 2(x + 1) = 0

(x + 1)(x - 2) = 0

x + 1 = 0 or x - 2 = 0

x = - 1 or x = 2