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Ask QuestionPosted by Aman Prajapati 6 years, 10 months ago
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Shweta Rao ?? 6 years, 10 months ago
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Sant Kumar 6 years, 10 months ago
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Kamlesh Jain 6 years, 10 months ago
Posted by Sakshi Bharti 6 years, 4 months ago
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Sia ? 6 years, 4 months ago
Let the original average speed of the train be x km/hr.
Time taken to cover 63 km {tex} = \frac{{63}}{x}{/tex} hours
Time taken to cover 72 km when the speed is increased by 6 km/hr {tex} = \frac{{72}}{{x + 6}}{/tex} hours
By the question,we have,
{tex}\frac{{63}}{x} + \frac{{72}}{{x + 6}} = 3{/tex}
{tex} \Rightarrow \frac{{21}}{x} + \frac{{24}}{{x + 6}} = 1{/tex}
{tex} \Rightarrow \frac{{21x + 126 + 24x}}{{{x^2} + 6x}} = 1{/tex}
{tex} \Rightarrow{/tex} 45x + 126 = x2 + 6x
{tex} \Rightarrow{/tex} x2 - 39x - 126 = 0
{tex} \Rightarrow{/tex} x2 - 42x + 3x - 126 = 0
{tex} \Rightarrow{/tex} x(x - 42) + 3(x - 42) = 0
{tex} \Rightarrow{/tex} (x - 42)(x + 3) = 0
{tex} \Rightarrow{/tex} x - 42 = 0 or x + 3 = 0
{tex} \Rightarrow{/tex} x = 42 or x = -3
Since the speed cannot be negative, {tex}x \ne -3{/tex}.
Thus, the original average speed of the train is 42 km/hr.
Posted by Anirudh Vijayan 6 years, 10 months ago
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Durgesh Kumar Singh 6 years, 10 months ago
Posted by Akshay Sonkamble 6 years, 10 months ago
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Posted by Sakshi Bharti 6 years, 10 months ago
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Divya Garg 6 years, 10 months ago
Posted by Abhishek Jain 6 years, 4 months ago
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Sia ? 6 years, 4 months ago
According to question{tex}{/tex}, the given equation is
x{tex}^2{/tex}(a2 + b2)+ 2x (ac + bd) +(c2 + d2)= 0 .
{tex}{/tex}Let D be the discriminant of this equation.
Therefore,D = 4(ac + bd)2 - 4(a2 + b2)(c2 + d2)
{tex} \Rightarrow{/tex} D = 4 [(ac+ bd)2 - (a2 + b2) (c2 + d2)]
{tex} \Rightarrow{/tex} D = 4 [a2c2 + b2d2 + 2ac.bd - a2c2 - a2d2 - b2c2 - b2d2]
{tex} \Rightarrow{/tex}D = 4 [2ac.bd - a2d2 - b2c2] = - 4[a2d2 + b2c2 - 2ad.bc] = -4(ad - bc){tex}^2{/tex}
It is given that ad {tex} \neq{/tex} bc.
{tex} \Rightarrow{/tex} ad - bc {tex} \neq{/tex} 0
{tex} \Rightarrow{/tex} (ad - bc){tex}^2{/tex} > 0
{tex} \Rightarrow{/tex} - 4 (ad - bc){tex}^2{/tex} < 0
{tex} \Rightarrow{/tex} D < 0.
Therefore, given equation has no real root.
Posted by Abhishek Jain 6 years, 4 months ago
- 1 answers
Sia ? 6 years, 4 months ago
According to the question,
{tex}\frac{m}{n}{x^2} + \frac{n}{m} = 1 - 2x{/tex}
{tex}\Rightarrow \frac{m}{n}{x^2} + 2x + \frac{n}{m} - 1 = 0{/tex}
{tex} \Rightarrow {x^2} + \frac{{2nx}}{m} + \frac{{{n^2}}}{{{m^2}}} - \frac{n}{m} = 0{/tex} [multiplying both sides by 'n' and dividing both sides by 'm']
{tex}\Rightarrow {x^2} + \frac{{2nx}}{m} + \frac{{{n^2} - mn}}{{{m^2}}} = 0{/tex}
To factorize {tex}{x^2} + \frac{{2nx}}{m} + \frac{{{n^2} - mn}}{{{m^2}}}{/tex}, we have to find two numbers {tex}'a'\ and\ 'b'{/tex} such that.
{tex}a + b = \frac{{2n}}{m}{/tex} and {tex}a b = \frac { n ^ { 2 } - m n } { m ^ { 2 } }{/tex}
Clearly, {tex}\frac{{n + \sqrt {mn} }}{m} + \frac{{n - \sqrt {mn} }}{m} = \frac{{2n}}{m}{/tex} and {tex}\frac{{\left( {n + \sqrt {mn} } \right)}}{m}\times\frac{{\left( {n - \sqrt {mn} } \right)}}{m} = \frac{{{n^2} - mn}}{{{m^2}}}{/tex} ({tex}\therefore a = \frac{{n + \sqrt {mn} }}{m}{/tex} and {tex}b = \frac{{n - \sqrt {mn} }}{m}{/tex})
{tex}\Rightarrow {x^2} + \frac{{2nx}}{m} + \frac{{{n^2} - mn}}{{{m^2}}} = 0{/tex}
{tex} \Rightarrow {x^2} + \frac{{\left( {n + \sqrt {mn} } \right)}}{m}x + \frac{{\left( {n - \sqrt {mn} } \right)}}{m}x{/tex}{tex} + \frac{{{n^2} - mn}}{{{m^2}}} = 0{/tex}
{tex} \Rightarrow x\left[ {x + \frac{{n + \sqrt {mn} }}{m}} \right] + \frac{{n - \sqrt {mn} }}{m}{/tex} {tex}\left[ {x + \frac{{n + \sqrt {mn} }}{m}} \right] = 0{/tex}
{tex} \Rightarrow \left( {x + \frac{{n - \sqrt {mn} }}{m}} \right)\left( {x + \frac{{n + \sqrt {mn} }}{m}} \right) = 0{/tex}
{tex} \Rightarrow x + \frac{{n - \sqrt {mn} }}{m} = 0{/tex} or {tex}x + \frac{{n + \sqrt {mn} }}{m} = 0{/tex}
{tex} \Rightarrow x = \frac{{- n - \sqrt {mn} }}{m}{/tex} or {tex}x = \frac{{ - n + \sqrt {mn} }}{m}{/tex}
Posted by Jot Virk?♠ 6 years, 10 months ago
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Jot Virk?♠ 6 years, 10 months ago
Aaradhya Anand 6 years, 10 months ago
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Diya Sharma 6 years, 10 months ago
1Thank You