Let $r$ be a root of the $x^{2} + 2 x + 6 = 0$ . The value of $(r + 2) (r + 3) (r + 4) (r + 5)$ is equal to.

51

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

- 51

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

- 126

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

126

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

Open in App

Solution

The correct option is B $- 126$
$r$ is a root of the quadratic eqution $x^{2} + 2 x + 6 = 0$
$\Rightarrow r^{2} + 2 r + 6 = 0$ ................(1)
Now $(r + 2) (r + 3) (r + 4) (r + 5)$
$= (r^{2} + 5 r + 6) (r^{2} + 9 r + 20)$
$= (3 r) (7 r + 14)$
$= 21 (r^{2} + 2 r)$
$= - 126$

Suggest Corrections

Similar questions

Let r be a root of the equation $x^{2}$ + 2x + 6 = 0. The value of (r + 2) (r + 3) (r + 4) (r + 5) is equal to.

Q. Let

r

be a root of the equation

x^{2} + 2 x + 6 = 0

. The value of

(r + 2) (r + 3) (r + 4) (r + 5)

is equal to.

Q. The greatest value of positive integer which divides

(r + 2) (r + 3) (r + 4) (r + 5) \forall r \in N

Find the value of:

p + q + 3r, when p = 1, q = 5, r = 2.

Let p and q be the roots of the equation $x^{2} - 2 x + A = 0,$ and let r and s be the roots of the equation $x^{2} - 18 x + B = 0.$ If p < q < r < s are in arithmetic progression. The value of (A+B) equals

Join BYJU'S Learning Program

Let r be a root of the x2+2x+6=0. The value of (r+2)(r+3)(r+4)(r+5) is equal to.

The correct option is B −126r is a root of the quadratic eqution x2+2x+6=0⇒r2+2r+6=0................(1)Now (r+2)(r+3)(r+4)(r+5) =(r2+5r+6)(r2+9r+20) =(3r)(7r+14) =21(r2+2r) =−126

Let $r$ be a root of the $x^{2} + 2 x + 6 = 0$ . The value of $(r + 2) (r + 3) (r + 4) (r + 5)$ is equal to.

The correct option is B $- 126$
$r$ is a root of the quadratic eqution $x^{2} + 2 x + 6 = 0$
$\Rightarrow r^{2} + 2 r + 6 = 0$ ................(1)
Now $(r + 2) (r + 3) (r + 4) (r + 5)$
$= (r^{2} + 5 r + 6) (r^{2} + 9 r + 20)$
$= (3 r) (7 r + 14)$
$= 21 (r^{2} + 2 r)$
$= - 126$