If 5a + |
|
= 5, the value of 9a2 + |
|
is |
3a | 25a2 |
If x = 2 + ?3, then the value of ?x + |
|
is : |
?x |
As we know from the algebra formula,
(a ? b)3 = a3 ? b 3 ? 3ab (a ? b) .......................(1)
As per given question,
a3 ? b 3 = 56 and a ? b = 2
Put these value in above equation (1) . We will get,
? 23 = 56 ? 3ab x 2
? 8 = 56 ? 3ab x 2
? 8 = 56 ? 6ab
? 6ab = 56 ? 8 = 48
? ab = 8 ...................... (i)
? a2 + b2 = (a ? b)2 + 2ab
? a2 + b2 = (a ? b)2 + 2ab
Put the value of
? a2 + b2 = 22 + 2 x 8
? a2 + b2 = 4 + 16 = 20
If (5x2 ? 3y2) : xy = 11 : 2, then the positive value of |
|
is : |
y |
Let the two numbers be P and Q.
According to given question,
Sum of two numbers = 24
? P + Q = 24 ...................... (1)
Product of two numbers = 143
and, PQ = 143 .......................... (2)
As we know the formula,
? P 2 + Q2 = (P + Q)2 ? 2PQ
Put the value from the equation (1) and (2), We will get
? P 2 + Q2 = (24)2 ? 2 × 143
? P 2 + Q2 = 576 ? 286 = 290
Given that :-
Now ,a3 + b3 + c3 = 3abc
( a+b+c )( a2 + b2 + c2 - ab - bc - ca ) + 3abc = 3abc
( a+b+c )( a2 + b2 + c2 - ab - bc - ca ) = 0
∴ a² + b² + c² - ab - bc - ca = 0 or a + b + c ≠ 0
Where ,
a = b = c
As indicated in the graph, the line passing through the points cuts Y-axis only.
Let the numbers are p and q.
p ? q = 3 ? (1)
p2 ? q2 = 39
(p ? q) (p + q) = 39 ? x + y = 13 ? (2)
Adding eqn (1) and (2)
p + q + p ? q = 16 ? p = 8
? q = 3
Hence, 8 is the larger number.
We know that , ?6 = 2.44, ?5 = 2.23, ?3 = 1.73
a = ?6 - ?5 = 2.44 - 2.23 = 0.21
b = ?5 - 2 = 2.23 - 2 = 0.23
c = 2 - ?3 = 2 - 1.73 = 0.27
Given that :- a2 + b2 + c2 = 2 (a ? b ? c) ? 3
? a2 + b2 + c2 - 2 (a ? b ? c) + 3 = 0
? a2 + b2 + c2 - 2a + 2b + 2c + 3 = 0
? ( a2 + 1 - 2a ) + ( b2 + 1 + 2b ) + ( c2 + 1 + 2c ) = 0
? ( a - 1 )2 + ( b + 1 )2 + ( c + 1 )2 = 0
This is possible when ( a - 1 )2 = 0, ( b + 1 )2 = 0, and ( c + 1 )2 = 0
? a = 1, b = ?1, c = ?1
Thus, 2a ? 3b + 4c = 2 (1) ? 3 (?1) + 4 (?1)
2a ? 3b + 4c = 2 + 3 ? 4 = 1.
Comments
There are no comments.Copyright ©CuriousTab. All rights reserved.