If one root of the equation ax^(2)+bx+c=0 is reciprocal of -Turito

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Question

If one root of the equation $a x squared plus b x plus c equals 0$ is reciprocal of the one of the roots of equation $a subscript 1 x squared plus b subscript 1 x plus c subscript 1 equals 0$ then

${(a a_{1} - c C_{1})}^{2} = (b c_{1} - b_{1} a) (b_{1} c - a_{1} b)$
${(a b_{1} - a_{1} b)}^{2} = (b c_{1} - b_{1} c) (c a_{1} - c_{1} a)$
${(b c_{1} - b_{1} c)}^{2} = (c a_{1} - a_{1} c) (a b_{1} - a_{1} b)$
${(a_{1} c - a c_{1})}^{2} = (b a_{1} - b_{1} a) (c_{1} b - b_{1} c)$

Hint:

In this question we will assume the root of the first equation so the root of the second equation will be its reciprocal, then we will substitute the value in the equation then we will solve both the equations to get the value of the root assumed. At last, we will put the value of the root in first equation and then solve it to get the required value.

The correct answer is: $open parentheses a a subscript 1 minus c C subscript 1 close parentheses squared equals open parentheses b c subscript 1 minus b subscript 1 a close parentheses open parentheses b subscript 1 c minus a subscript 1 b close parentheses$

Let one of the roots of the equation $a x squared plus b x plus c equals 0$ be $alpha$ .
Then as per question one of the roots of the equation $a subscript 1 x squared plus b subscript 1 x plus c subscript 1 equals 0$ will be $1 over alpha$ .
$a x squared plus b x plus c equals 0 rightwards double arrow a alpha squared plus b alpha plus c equals 0 space _ _ _ _ _ _ _ _ _ e q u a t i o n space 1$
$a subscript 1 x squared plus b subscript 1 x plus c subscript 1 equals 0 rightwards double arrow a subscript 1 open parentheses 1 over alpha close parentheses squared plus b subscript 1 open parentheses 1 over alpha close parentheses plus c subscript 1 equals 0 _ _ _ _ _ _ _ _ _ _ _ e q u a t i o n space 2$
Multiplying equation 1 by $c subscript 1$ and equation 2 by a, then subtracting, we get:
$stack attributes charalign center stackalign right end attributes row a c subscript 1 alpha squared plus b c subscript 1 alpha plus c c subscript 1 equals 0 end row row a c subscript 1 alpha squared plus a b subscript 1 alpha plus a subscript 1 a equals 0 end row row minus none end row horizontal line row b c subscript 1 alpha minus a b subscript 1 alpha plus c c subscript 1 minus a subscript 1 a equals 0 end row end stack$
$rightwards double arrow alpha open parentheses b c subscript 1 minus b subscript 1 a close parentheses plus c c subscript 1 minus a subscript 1 a equals 0 rightwards double arrow alpha equals fraction numerator a subscript 1 a minus c c subscript 1 over denominator b c subscript 1 minus a b subscript 1 end fraction$
putting the value in equation 1
$a open parentheses fraction numerator a a subscript 1 minus c c subscript 1 over denominator b c subscript 1 minus a b subscript 1 end fraction close parentheses squared plus b open parentheses fraction numerator a a subscript 1 minus c c subscript 1 over denominator b c subscript 1 minus a b subscript 1 end fraction close parentheses plus c equals 0 rightwards double arrow a open parentheses a a subscript 1 minus c c subscript 1 close parentheses squared plus b open parentheses a a subscript 1 minus c c subscript 1 close parentheses open parentheses b c subscript 1 minus a b subscript 1 close parentheses plus c open parentheses b c subscript 1 minus a b subscript 1 close parentheses squared equals 0 rightwards double arrow a open parentheses a a subscript 1 minus c c subscript 1 close parentheses squared plus open parentheses b c subscript 1 minus a b subscript 1 close parentheses open parentheses a a subscript 1 b minus c c subscript 1 b plus a b subscript 1 c plus b c c subscript 1 close parentheses equals 0 rightwards double arrow a open parentheses a a subscript 1 minus c c subscript 1 close parentheses squared plus open parentheses b c subscript 1 minus a b subscript 1 close parentheses a left parenthesis a subscript 1 b minus b subscript 1 c right parenthesis equals 0 rightwards double arrow a open parentheses a a subscript 1 minus c c subscript 1 close parentheses squared equals negative open parentheses b c subscript 1 minus a b subscript 1 close parentheses a left parenthesis a subscript 1 b minus b subscript 1 c right parenthesis rightwards double arrow open parentheses a a subscript 1 minus c c subscript 1 close parentheses squared equals open parentheses b c subscript 1 minus a b subscript 1 close parentheses left parenthesis b subscript 1 c minus a subscript 1 b right parenthesis$

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Question

If one root of the equation $a x squared plus b x plus c equals 0$ is reciprocal of the one of the roots of equation $a subscript 1 x squared plus b subscript 1 x plus c subscript 1 equals 0$ then

${(a a_{1} - c C_{1})}^{2} = (b c_{1} - b_{1} a) (b_{1} c - a_{1} b)$
${(a b_{1} - a_{1} b)}^{2} = (b c_{1} - b_{1} c) (c a_{1} - c_{1} a)$
${(b c_{1} - b_{1} c)}^{2} = (c a_{1} - a_{1} c) (a b_{1} - a_{1} b)$
${(a_{1} c - a c_{1})}^{2} = (b a_{1} - b_{1} a) (c_{1} b - b_{1} c)$

Hint:

Answer

The correct answer is: $open parentheses a a subscript 1 minus c C subscript 1 close parentheses squared equals open parentheses b c subscript 1 minus b subscript 1 a close parentheses open parentheses b subscript 1 c minus a subscript 1 b close parentheses$

Solution

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Question

If one root of the equation is reciprocal of the one of the roots of equation then

Hint:

Answer

The correct answer is:

Solution

Let one of the roots of the equation be .Then as per question one of the roots of the equation will be.Multiplying equation 1 by and equation 2 by a, then subtracting, we get:putting the value in equation 1

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If one root of the equation $a x squared plus b x plus c equals 0$ is reciprocal of the one of the roots of equation $a subscript 1 x squared plus b subscript 1 x plus c subscript 1 equals 0$ then

The correct answer is: $open parentheses a a subscript 1 minus c C subscript 1 close parentheses squared equals open parentheses b c subscript 1 minus b subscript 1 a close parentheses open parentheses b subscript 1 c minus a subscript 1 b close parentheses$