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Ask QuestionPosted by Piyush Bhumiputra 6 years, 4 months ago
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Shaik Haneef 6 years, 4 months ago
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Sia ? 6 years, 4 months ago
Collinearity of a set of points is the property of their lying on a single line.
Rishu Kumar 6 years, 4 months ago
Posted by Niyati Chauhan 6 years, 4 months ago
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Gaurav Seth 6 years, 4 months ago
p(x) = 4u²+8u
4u ( u + 2 ) = 0
[4u + 0 ] [ u + 2 ] ------> factorized form
hence ,
u = 0
u = -2
Here ,
a = 4
b = 8
c = 0
Relation between zeros and its coefficients :-
Sum of zeroes = 0 + -2 = - 2
= -b/a = -8/4 = -2
Product of zeroes = 0 x -2 = 0
= c/a = 0/4 = 0
Posted by Ghanshyam Sisode 6 years, 4 months ago
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Prashant Thakur 6 years, 4 months ago
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Posted by Sahitya Pandey 6 years, 4 months ago
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Gaurav Seth 6 years, 4 months ago
√3x^2+11x+6√3
By splitting middle term
√3x^2+9x+2x+6√3
Taking common
√3x(x+3√3)+2(x+3√3)
(√3x+2) (x+3√3)
Posted by Shravya Shetty 6 years, 4 months ago
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Sia ? 6 years, 4 months ago
We have, abx2 + (b2 -ac) x-bc = 0
{tex}\implies{/tex}abx2 + b2 x - acx - bc = 0
{tex}\implies{/tex}bx ( ax+b) - c (ax + b) = 0
{tex}\implies{/tex}(ax + b) (bx - c) = 0
Either ax+b = 0 or bx - c = 0
{tex}\implies x = -{b \over a},\, {c \over b}{/tex}
Hence, {tex}x = -{b \over a},\, {c \over b}{/tex} are the required solutions.
Posted by Abhay Rajput 6 years, 4 months ago
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Posted by Harshit Sen 6 years, 4 months ago
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Posted by Piyush Singh 6 years, 4 months ago
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Sia ? 6 years, 4 months ago
We have (x + 2)3 = 2x (x2 - 1)
{tex}\implies{/tex}x3 + 6x2 + 12x + 8 = 2x3 - 2x
{tex}\implies{/tex}x3 - 6x2 - 14x - 8 = 0
it is not of the form ax2 + bx + c = 0, {tex}a \ne 0{/tex}
{tex}\therefore{/tex} the given equation is not a quadratic equation.
Posted by Bicram Gharami 6 years, 4 months ago
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#Aditi~ Angel???? 6 years, 4 months ago
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Prashant Saini 6 years, 4 months ago
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Mayank Sharma 6 years, 4 months ago
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Ayush Modi 6 years, 4 months ago
Posted by Sreenandana M . S 6 years, 4 months ago
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Gaurav Seth 6 years, 4 months ago
Let us suppose that the numbers (a+b)/2 and (a-b)/2 are either both odd or both even.
Case1) When (a+b)/2 and (a-b)/2 are odd.
We know that the sum or difference of two odd numbers is even, hence the sum of the numbers (a+b)/2 and (a-b)/2 must be even.
So, (a+b)/2 +(a-b)/2=a must be even which is not correct as we are given that a is odd positive integer.
(If we take the difference, we will get the value as equal to b).
This leads to a contradiction. Hence (a+b)/2 +(a-b)/2 cannot be both odd.
Case 2) When (a+b)/2 and (a-b)/2 are even.
Again, the sum or difference of two even numbers is even.
So, (a+b)/2 +(a-b)/2=a must be even, which is not correct as we are given that a is odd positive integer. (If we take the difference, we will get value equal to b). This again leads to a contradiction. Hence (a+b)/2 +(a-b)/2 cannot be both even.
So, the given two numbers cannot be both even or both odd. Hence, there is only one possibility that one out of a+b/2 and a-b/2 is odd and the other is even.
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Piyush Bhumiputra 6 years, 3 months ago
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