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If x=−9 is a root of [x372x276x]=0,   then the other two roots are ____.

0 2 Views | Posted 3 months ago
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  • V

    Answered by

    Vishal Baghel | Contributor-Level 10

    3 months ago

    This is a  Fill in the blanks as classified in NCERT Exemplar

    Sol:

    W e     h a v e ,     | x 3 7 2 x 2 7 6 x | = 0 E x p a n d i n g     a l o n g     R 1 ⇒                                                 x | x 2 6 x | − 3 | 2 2 7 x | + 7 | 2 x 7 6 | = 0 ⇒         x ( x 2 − 1 2 ) − 3 ( 2 x − 1 4 ) + 7 ( 1 2 − 7 x ) = 0 ⇒                                           x 3 − 1 2 x − 6 x + 4 2 + 8 4 − 4 9 x = 0 ⇒                                                                                                         x 3 − 6 7 x + 1 2 6 = 0                                                 … ( 1 ) T h e     r o o t s     o f     t h e     e q u a t i o n     m a y     b e     t h e     f a c t o r s     o f     1 2 6     i . e . ,     2 × 7 × 9 9  is  the  given  root  of  the  determinant  put  x=2  in  eq.(1)                                                                       ( 2 ) 3 − 6 7 × 2 + 1 2 6               ⇒               8 − 1 3 4 + 1 2 6 = 0 H e n c e ,     x = 2     i s     t h e     o t h e r     r o o t . N o w ,     p u t     x = 7     i n     e q . ( 1 )                                                                       ( 7 ) 3 − 6 7 × 7 + 1 2 6               ⇒               3 4 3 − 4 6 9 + 1 2 6 = 0 Hence,  x=7  is  also  the  other  root  of  the  determinant.

Similar Questions for you

A = [ 1 0 0 0 α β 0 β α ] then α is (if α, β Î I)
A
alok kumar singh

|2A| = 27

8|A| = 27

Now |A| = α2–β2 = 24

α2 = 16 + β2

α2– β2 = 16

(α–β) (α+β) = 16

->α + β = 8 and

α – β = 2

->α = 5 and β = 3

If A = [ 2 1 − 1 2 ] , B = [ 1 0 1 1 ] , C = ABAT and X = AC2AT then |X| is equal to
A
alok kumar singh

|A| = 3

|B| = 1

->|C| = |ABAT| = |A|B|A7| = |A|2|B|

= 9

->|X| = |A|C|2|AT|

= 3 * 92 * 3 = 9 * 92 = 729

Matrix A of order 3 × 3 is such that |A| = 2 if n = | a d j ( a d j ( a d j . . . . ( a ) ) ) | ︸ 2 0 2 4   t i m e s  then remainder when n is divided by 9 is
A
alok kumar singh

|A| = 2

a d j ( a d j ( a d j . . . . ( a ) ) ) ? 2 0 2 4   t i m e s = | A | ( n − 1 ) 2 0 2 4            

= | A | ( n − 1 ) 2 0 2 4                                             

= 2 2 2 0 2 4                                          

2 2 0 2 4 = ( 2 2 ) 2 2 0 2 2 = 4 ( 8 ) 6 7 4 = 4 ( 9 − 1 ) 6 7 4             

-> 2 2 0 2 4 ≡ 4 ( m o d   9 )

->,  2 2 0 2 4 ≡ 9 m + 4  m ¬ even

2 9 m + 4 ≡ 1 6 ⋅ ( 2 3 ) 3 m ≡ 1 6 ( m o d   9 )

7

If ∫ 0 2 ( 2 x − 2 x − x 2 ) d x = ∫ 0 1 ( 1 − 1 − y 2 − y 2 2 ) d y + ∫ 1 2 ( 2 − y 2 2 ) d y + I     t h e n     I     e q u a l s  
V
Vishal Baghel

∫ 0 2 2 x d x − ∫ 0 2 2 x − x 2 d x = ∫ 0 1 d y − ∫ 0 1 1 − y 2 d y − ∫ 0 2 y 2 2 d y + ∫ 1 2 2 d y + I

⇒ 8 3 − ∫ 0 1 1 − t 2 d t = 1 − 8 6 + 2 + I

I = 1 − ∫ 0 1 1 − t 2 d t

If  ∫ 0 2 ( 2 x − 2 x − x 2 ) d x = ∫ 0 1 ( 1 − 1 − y 2 − y 2 2 ) d y + ∫ 1 2 ( 2 − y 2 2 ) d y + I     t h e n     I     e q u a l s

A
alok kumar singh

∫ 0 2 2 x d x − ∫ 0 2 2 x − x 2 d x = ∫ 0 1 d y − ∫ 0 1 1 − y 2 d y − ∫ 0 2 y 2 2 d y + ∫ 1 2 2 d y + I

⇒ 8 3 − ∫ 0 1 1 − t 2 d t = 1 − 8 6 + 2 + I

I = 1 − ∫ 0 1 1 − t 2 d t

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