You visited us 1 times! Enjoying our articles? Unlock Full Access!

nCr Definitions and Properties

Let r be a ro...

Question

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.

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 C
-126

r be a root $\Rightarrow r^{2}$ + 2r + 6 = 0 ...................(1)

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

= ( $r^{2}$ + 5r +6) ( $r^{2}$ + 9r + 20)

= (3r) (7r + 14) using (i)

= 21 ( $r^{2}$ + 2r)

= -126 using (i)

Suggest Corrections

Similar questions

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. 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.

Q. The greatest value of positive integer which divides

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

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

Q. Let

p

and

q

be 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, then find the value of

A + B

Join BYJU'S Learning Program

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

The correct option is C -126 r be a root ⇒r2 + 2r + 6 = 0 ...................(1) now (r+2) (r+3) (r+4) (r+5) = (r2 + 5r +6) (r2+ 9r + 20) = (3r) (7r + 14) using (i) = 21 (r2+ 2r) = -126 using (i)

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.

The correct option is C
-126

r be a root $\Rightarrow r^{2}$ + 2r + 6 = 0 ...................(1)

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

= ( $r^{2}$ + 5r +6) ( $r^{2}$ + 9r + 20)

= (3r) (7r + 14) using (i)

= 21 ( $r^{2}$ + 2r)

= -126 using (i)