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Ask QuestionPosted by Deep Sidhu 6 years, 4 months ago
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Posted by Pooja Mania 6 years, 4 months ago
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Posted by Jai Shree Shyam 6 years, 3 months ago
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Sia ? 6 years, 3 months ago
Taking {tex}\frac { 1 } { x } = u{/tex} and {tex}\frac { 1 } { y } = v{/tex}, the above system of equation becomes
{tex}{/tex}au - vb = 0............ (i)
{tex}{/tex} {tex}{/tex}ab2u + a2bv = a2 + b2.......... (ii)
By cross-multiplication, using (i) and (ii) we have
{tex}\frac { u } { - b \times - \left( a ^ { 2 } + b ^ { 2 } \right) - a ^ { 2 } b \times 0 } = \frac { - v } { a \times - \left( a ^ { 2 } + b ^ { 2 } \right) - a b ^ { 2 } \times 0 } = \frac { 1 } { a \times a ^ { 2 } b - a b ^ { 2 }( - b) }{/tex}
{tex}\Rightarrow \quad \frac { u } { b \left( a ^ { 2 } + b ^ { 2 } \right) } = \frac { - v } { - a \left( a ^ { 2 } + b ^ { 2 } \right) } = \frac { 1 } { a ^ { 3 } b + a b ^ { 3 } }{/tex}
{tex}\Rightarrow \quad \frac { u } { b \left( a ^ { 2 } + b ^ { 2 } \right) } = \frac { v } { a \left( a ^ { 2 } + b ^ { 2 } \right) } = \frac { 1 } { a b \left( a ^ { 2 } + b ^ { 2 } \right) }{/tex}
{tex}\Rightarrow \quad u = \frac { b \left( a ^ { 2 } + b ^ { 2 } \right) } { a b \left( a ^ { 2 } + b ^ { 2 } \right) } = \frac { 1 } { a } \text { and } v = \frac { a \left( a ^ { 2 } + b ^ { 2 } \right) } { a b \left( a ^ { 2 } + b ^ { 2 } \right) } = \frac { 1 } { b }{/tex}
Now, {tex}u = \frac { 1 } { a } {/tex}
{tex}\Rightarrow \frac { 1 } { x } = \frac { 1 } { a } {/tex}
{tex}\Rightarrow {/tex} x = a
and {tex}v = \frac { 1 } { b } {/tex}
{tex}\Rightarrow \frac { 1 } { y } = \frac { 1 } { b }{/tex}
{tex} \Rightarrow{/tex} y = b
Hence, the solution of the given system of equation is x = a, y = b.
Posted by Darsh Patel 6 years, 4 months ago
- 2 answers
Posted by Md Wasim Akram 6 years, 4 months ago
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Posted by Hariom Yaduvanshi 6 years, 4 months ago
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Nm ???? 6 years, 4 months ago
Priyanshu Rikhari 6 years, 4 months ago
Posted by Nee Ti 6 years, 3 months ago
- 2 answers
Sia ? 6 years, 3 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 Smarty Boy 3 6 years, 4 months ago
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Posted by Raju Tulasigeri 6 years, 4 months ago
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Posted by A I 6 years, 4 months ago
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Posted by Mahi Garg 6 years, 4 months ago
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Sia ? 6 years, 4 months ago
Zero of a polynomial is a solution to the polynomial equation, P(x) = 0 but Root of a polynomial is that value of x that makes the polynomial equal to 0.
Iron Man 6 years, 4 months ago
Posted by Chinmaya Senapati 6 years, 4 months ago
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Sia ? 6 years, 4 months ago
We will solve this by contradiction method i.e.,
Assume {tex}5 - \sqrt { 3 } = \frac { p } { q }{/tex} be a rational number.
{tex}\therefore 5 - \frac { p } { q } = \sqrt { 3 }{/tex} or {tex}\frac { 5 q - p } { q } = \sqrt { 3 }{/tex},
Since p,q are integers, therefore {tex}\frac { 5 q - p } { q }{/tex} is a rational number, which is a contradiction,since {tex}\sqrt 3{/tex} is an irrational number.
Therefore, our supposition is wrong and hence, 5 - {tex}\sqrt 3{/tex} is irrational.
Posted by Norgyal Bhutia 6 years, 4 months ago
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Posted by Prakhar Binjwa 6 years, 4 months ago
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Posted by Mohd Ayan 6 years, 4 months ago
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Posted by Mohd Ayan 6 years, 4 months ago
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Sandeep Yadav 6 years, 4 months ago
Student Lyf ? 6 years, 4 months ago
Posted by Lokanath Lokanath 6 years, 4 months ago
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Surya Narayan Swain 6 years, 4 months ago
Posted by Ritish Gangwar 6 years, 4 months ago
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Posted by Tanya Kumari 6 years, 4 months ago
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Karpagam Selvakumar 6 years, 4 months ago
Posted by Naruto Uzumaki??? 6 years, 4 months ago
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Yogita Ingle 6 years, 4 months ago
38220>196 we always divide greater number with smaller one.
Divide 38220 by 196 then we get quotient 195 and no remainder so we can write it as
38220 = 196 * 195 + 0
As there is no remainder so deviser 196 is our HCF
Posted by Arnav . 6 years, 4 months ago
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Posted by Mohd Ayan 6 years, 4 months ago
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Posted by Deepu ?? 6 years, 4 months ago
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Posted by Avtar Singh 6 years, 4 months ago
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Posted by Karan Kumar 6 years, 4 months ago
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Yogita Ingle 6 years, 4 months ago
A polynomial equation whose degree is 2, is known as quadratic equation. A quadratic equation in its standard form is represented as: ax2 + bx + c = 0 , where a, b and c are real numbers such that a ≠ 0 is a variable.
The number of roots of a polynomial equation is equal to its degree. So, a quadratic equation has two roots.
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Aadya Singh 6 years, 4 months ago
1Thank You