If $(y^{2} + 5 y) (y^{2} + 5 y - 2) - 24 = 0$ , then $y = 1, - 6, - 1, - 4$ .

Yes

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

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

Ambiguous

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

Data insufficient

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

Open in App

Solution

The correct option is A Yes
$(y^{2} + 5 y) (y^{2} + 5 y - 2) - 24 = 0$
Let $y^{2} + 5 y = m$
Hence, reframing the equation,
$m (m - 2) - 24 = 0$
$m^{2} - 2 m - 24 = 0$
$m^{2} - 6 m + 4 m - 24 = 0$
$(m - 6) (m + 4) = 0$
$m = - 4, 6$
Now, when $m = - 4$
$y^{2} + 5 y = - 4$
$y^{2} + 5 y + 4 = 0$
$y = - 1, - 4$
Now, when $m = 6$
$y^{2} + 5 y = 6$
$y^{2} + 6 y - y - 6 = 0$
$y = - 6, 1$
Hence, $y = 1, - 1, - 4. - 6$

Suggest Corrections

Similar questions

(y^{2} + 5 y) (y^{2} + 5 y - 2) - 24 = 0

-

-

(y^{2} + 5 y) (y^{2} + 5 y - 2) - 24

If \frac{4 x - 5 y}{3 x + y} = 1, then \frac{2 x^{2} + y^{2}}{2 x^{2} - y^{2}} =

y = 4, - 2, - 3

y^{2} + 5 y + 6 = 0

Join BYJU'S Learning Program