Question

1)
Solve the following equations. (x2 + x) (x2 + x - 2) = 24
the following quadratic from th​

Answer 1

${\huge {\mathfrak {Answer :-}}}$

____________

(x^2 + x) (x^2 + x - 2) = 24

Let, y = x^2 + x

✔️ So, y (y - 2) = 24

Or, y^2 - 2y - 24 = 0

Or, y^2 - 6y + 4y - 24 = 0

Or, y(y - 6) + 4(y - 6) = 0

Or, (y - 6)(y + 4) = 0.

Or, y = 6, - 4 ✔️

Thus, x^2 + x = 6, - 4

✔️ So, x^2 + x = 6

Or, x^2 + x - 6 = 0

Or, x^2 + 3x - 2x - 6 = 0

Or, x(x + 3) - 2(x + 3) = 0

Or, (x + 3)(x - 2) = 0

Or, x = 2, - 3 ✔️

And, x^2 + x = - 4

Or, x^2 + x + 4 = 0

But, D = b^2 - 4ac

= (1)^2 - 4.1. 4

= 1 - 16 = - 15

Since, D < 0, so it is not possible to have such a quadratic equation.

Hence, x = 2, - 3.

____________

Thanks..

Answer 2

Step-by-step explanation:

(x^2 + x) (x^2 + x - 2) = 24

Let, y = x^2 + x

✔️ So, y (y - 2) = 24

Or, y^2 - 2y - 24 = 0

Or, y^2 - 6y + 4y - 24 = 0

Or, y(y - 6) + 4(y - 6) = 0

Or, (y - 6)(y + 4) = 0.

Or, y = 6, - 4 ✔️

Thus, x^2 + x = 6, - 4

✔️ So, x^2 + x = 6

Or, x^2 + x - 6 = 0

Or, x^2 + 3x - 2x - 6 = 0

Or, x(x + 3) - 2(x + 3) = 0

Or, (x + 3)(x - 2) = 0

Or, x = 2, - 3 ✔️

And, x^2 + x = - 4

Or, x^2 + x + 4 = 0

But, D = b^2 - 4ac

= (1)^2 - 4.1. 4

= 1 - 16 = - 15

Since, D < 0, so it is not possible to have such a quadratic equation.

Hence, x = 2, - 3.