Find the roots of the following equation:
$(x^{2} + 2 x) (x^{2} + 2 x - 11) + 24 = 0$ , then

x = 4, 2, - 3, 1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

x = - 4, - 2, - 3, 1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

x = - 4, 2, - 3, - 1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

x = - 4, 2, - 3, 1

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is A $x = - 4, 2, - 3, 1$
Given equation is,
$(x^{2} + 2 x) (x^{2} + 2 x - 11) + 24 = 0$
Let $x^{2} + 2 x = y$ ............(1)
then, $(y) (y - 11) + 24 = 0$
$y^{2} - 11 y + 24 = 0$
$y^{2} - 8 y - 3 y + 24 = 0$
$∴ y = 3, 8$
Now substituting value of $y$ in equation (1), we get
$x^{2} + 2 x = 3$ and $x^{2} + 2 x = 8$
$x^{2} + 3 x - x - 3 = 0$ and $x^{2} + 4 x - 2 x - 8 = 0$
$(x + 3) (x - 1) = 0$ and $(x + 4) (x - 2) = 0$
$∴ x = 1, 2, - 3, - 4$

Suggest Corrections

Similar questions

(1 + α) x + β y + z = 2, α x + (1 + β) y + z = 3, α x + β y + 2 z = 2

(α, β)

The points (5, -4, 2) (4, -3, 1), (7, 6, 4) and (8, -7, 5) are the vertices of

A = {1, 2}

B = {2, 3, 4}

B \times A = {(2, 1), (2, 2), (3, 1), (3, 2), (4, 1), (4, 2)}

1

0

The value/s of x when (x - 4) (3x + 2) = 0 ________

A \times B = {(1, 1), (1, 2), (1, 3), (3, 1), (3, 2), (3, 3), (4, 1), (4, 2), (4, 3)}

Join BYJU'S Learning Program