Question

(x²+x)²-(x²+x)-2=o quadratic equation
solve it by reducing it to quadratic equation
​

Answer 1

Step-by-step explanation:

Let (x² + x) be a.

So, the equation becomes,

a² - a - 2 = 0

a² - (2a - a) - 2 = 0

a² + a - 2a - 2 = 0

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

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

a + 1 = 0 or a - 2 = 0

a = -1 or a = 2

x² + x = -1 or x² + x = 2

x² + x + 1 = 0 or x² + x - 2 = 0

(here, find x using

quadratic formula)

x² + 2x - x - 2 = 0

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

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

x = 1, -2

Answer 2

Answer:

x =  -2,1,$\frac{-1 + i\sqrt{3} }{2}$,$\frac{-1 - i\sqrt{3} }{2}$

Step-by-step explanation:

(x²+x)²-(x²+x)-2= 0

let X = (x²+x)

Now the equation becomes,

X² - X - 2 = 0

=> X² - 2X + X - 2 = 0

=> X(X - 2) + X - 2 = 0

=> (X - 2)(X + 1) = 0

=> (X - 2) = 0 or (X +1)=0

=> x²+x-2 = 0  or x²+x+1 = 0

take x²+x-2 = 0,

=> x²+2x-x-2 = 0

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

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

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

=> x= - 2 or x = 1

take x²+x+1 = 0

using formula,

x = $\frac{-1 \pm \sqrt{1^2 - 4(1)(1)} }{2(1)}$

  = $\frac{-1 \pm \sqrt{1 - 4} }{2}$

  = $\frac{-1 \pm \sqrt{-3} }{2}$

  = $\frac{-1 \pm i\sqrt{3} }{2}$

  = $\frac{-1 + i\sqrt{3} }{2}$ or $\frac{-1 - i\sqrt{3} }{2}$

so, the solutions are, -2,1,$\frac{-1 + i\sqrt{3} }{2}$,$\frac{-1 - i\sqrt{3} }{2}$