If a root of the equation $a_{1} x^{2} + b_{1} x + c_{1} = 0$ is the reciprocal of a root of the equation $a_{2} x^{2} + b_{2} x + c_{2} = 0$ then

(a_{1} a_{2} - c_{1} c_{2})^{2} = (a_{1} b_{2} - b_{1} c_{2}) (a_{2} b_{1} - b_{2} c_{1})

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(a_{1} a_{2} - b_{1} b_{2})^{2} = (a_{1} b_{2} - b_{1} c_{2}) (a_{2} b_{1} - b_{2} c_{1})

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(b_{1} c_{2} - b_{2} c_{1})^{2} = (a_{1} b_{2} - b_{1} c_{2}) (a_{2} b_{1} - b_{2} c_{1})

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(b_{1} c_{2} - b_{2} c_{1})^{2} = (a_{1} b_{2} + b_{1} c_{2}) (a_{2} b_{1} - b_{2} c_{1})

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Solution

The correct option is A $(a_{1} a_{2} - c_{1} c_{2})^{2} = (a_{1} b_{2} - b_{1} c_{2}) (a_{2} b_{1} - b_{2} c_{1})$

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Similar questions

Q. If the equations

a_{1} x^{2} + b_{1} x + c_{1} = 0

and

a_{2} x^{2} + b_{2} x + c_{2} = 0

have a common root, then the value of

(a_{1} b_{2} - a_{2} b_{1}) (b_{1} c_{2} - b_{2} c_{1})

For the given pair of linear equations $a_{1} x + b_{1} y + c_{1} = 0$ and $a_{2} x + b_{2} y + c_{2} = 0,$ the solution $(x, y)$ is

The solution (x,y) for the given pair of linear equations is $a_{1}$ x + $b_{1}$ y + $c_{1}$ =0 and $a_{2}$ x + $b_{2}$ y + $c_{2}$ =0

Solve the two equations using method of elimination.

$a_{1} x + b_{1} y + c_{1} = 0$
$a_{2} x + b_{2} y + c_{2} = 0$

In cross multiplication method of solving
$a_{1} x + b_{1} y + c_{1} = 0 a n d a_{2} x + b_{2} y + c_{2} = 0, x = \frac{b_{1} c_{2} - b_{2} c_{1}}{a_{1} b_{2} - a_{2} b_{1}}$

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If a root of the equation a1x2+b1x+c1=0 is the reciprocal of a root of the equation a2x2+b2x+c2=0 then

The correct option is A (a1a2−c1c2)2=(a1b2−b1c2)(a2b1−b2c1)

If a root of the equation $a_{1} x^{2} + b_{1} x + c_{1} = 0$ is the reciprocal of a root of the equation $a_{2} x^{2} + b_{2} x + c_{2} = 0$ then

The correct option is A $(a_{1} a_{2} - c_{1} c_{2})^{2} = (a_{1} b_{2} - b_{1} c_{2}) (a_{2} b_{1} - b_{2} c_{1})$