Question

(x^2+x)^2-(x^2+x)-2=0 solve by quardratic equation

Answer 1

Answer:

x = -2, 1

Step-by-step explanation:

In the given equation,

$(x^{2}+x)^{2}-(x^{2}+x)-2=0$

Here,

We can directly put the value of, $(x^{2}+x)$ as t.

So,

The equation simplifies to,

$(t)^{2}-t-2=0\\So,\\t^{2}-2t+t-2=0\\t(t-2)+1(t-2)=0\\(t+1)(t-2)=0\\t=-1,2$

Therefore, the value of t is equal to -1 and 2.

Now,

$x^{2}+x=t\\So,\\x^{2}+x=-1\\x^{2}+x+1=0$

As the value of the discriminant is less than 0.

This equation will give imaginary values of x.

Now,

when t = 2.

$x^{2}+x-2=0\\x^{2}+2x-x-2=0\\x(x+2)-1(x+2)=0\\(x-1)(x+2)=0\\x=-2,1$

Therefore, the value of x is given by -2 and 1.