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What should be subtracted to x^{2}+2x+2 to make it a perfect square?
By subtracting 1 from 2,you will get 1 that makes the whole square of (x+1)=1
If a + b = 12 and a^{2} + b^{2} = 60, the value of a^{3} + b^{3} is:
The correct option is Option C.
(a + b)² = a² + b² + 2ab
(12)² = 60 + 2ab
ab = (144  60)/2
ab = 84/2
ab = 42
Now,
(a + b)³ = a³ + b³ + 3ab(a+b)
(12)³ = a³ + b³ + (3 × 42 × 12)
a³ + b³ = 1728  1512
Therefore, a³ + b³ = 216
One of the factors of (16y^{2} – 1) + (1 4y)^{2} is
(1001)²= (100)²+(1)²+2×100×1 10000+1+200 =10201
= ( a² + b² + c²  ab  bc  ca )
By multiplying it by 2 and dividing it by 2.
= 2 ( a² + b² + c²  ab  bc  ca ) ÷ 2
= ( 2a² + 2b² + 2c²  2ab  2bc  2ca ) ÷ 2
= ( a² + a² + b² + b² + c² + c²  2ab  2bc  2ca ) ÷ 2
= ( a² + b²  2ab + b² + c²  2bc + a² + c²  2ca ) ÷ 2
= [ ( a  b )² + ( b  c )² + ( a  c )² ] ÷ 2
Now , whatever is the value of ( a  b ) , ( b  c ) and ( a  c ) but its square will be always positive, and 2 is also a positive number.
So, the sum of ( a  b )², ( b  c )² and ( a  c )² will be a positive number and if a positive number is divided by a positive number then the result is also a positive number.
So, it is a positive number,hence it can't be a negative number.
(100+2)×(1002). [(a+b)(ab)=a^2b^2
100^2  2^2
10000  4
9996
If a  b = 3 and a^{2} + b^{2} = 29, find the value of ab.
2ab = (a^{2} + b^{2})  (a  b)^{2}
= 29  9 = 20
⇒ ab = 10.
What should be added to x^{2}+2x+0.5 to make it a perfect square?
Let p(x) = x^{2} + 2x + 0.5
To find what must be added to p(x) to make it perfect square.
if we add 0.5 to p(x) we will get
p(x) = x2 + 2x + 0.5 + 0.5
p(x) = x2 + 2x + 1
p(x) = x2 + 1x + 1x + 1
p(x) = x(x + 1) + 1(x + 1)
p(x) = (x+1)(x+1)
p(x) = (x+1)^{2}
which is a perfect square.
Therefore 0.5 must be added to the given expression to make it perfect square.
The missing figure in the expression
(5p +…)^{2} = ….+ 40 pq + 16q^{2} are:
(100+1)×(100  1) it is the formula of a²b² so, (100)² (1)² = 100001 = 9999.
If 2x + 3y = 12 and xy = 5, find the value of 4x^{2} + 9y^{2}.
The expansion of (2a  3b + 5c)² : (2a)² + ( 3b)² + (5c)² + 2× 2a × (3b) + 2× (3b)× 5c + 2× 5c + 2a. = 4a² + 9b² + 25c²  12ab  30bc + 20ac
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