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If $ad
eq bc$, then prove that the equation $\left( a^{2} +b^{2}\right) x^{2} +2( ac\ +\ bd) \ x+\left( c^{2} +d^{2}\right) =0$ has no real roots.
Given: The equation $( a^{2} +b^{2}) x^{2} +2( ac\ +\ bd) \ x+( c^{2} +d^{2}) =0$, ad$\
eq $bc,
eq $bc,
To do: To prove that the given equation has no real roots.
Solution:
Given $ad\
eq bc$, for the equation $\left( a^{2} +b^{2}\right) x^{2} +2( ac\ +\ bd) \ x+\left( c^{2} +d^{2}\right) =0$
eq bc$, for the equation $\left( a^{2} +b^{2}\right) x^{2} +2( ac\ +\ bd) \ x+\left( c^{2} +d^{2}\right) =0$
For this equation not to have real roots its discriminant $D< 0$
$D=4( ac\ +\ bd)^{2} -4\left( a^{2} +b^{2}\right)\left( c^{2} +d^{2}\right)$
$D=4a^{2} c^{2} +4b^{2} d^{2} +8abcd-4a^{2} c^{2} -4a^{2} d^{2} -4b^{2} c^{2} -4b^{2} d^{2}$
$D=-4\left( a^{2} d^{2} +b^{2} c^{2} -2abcd\right)$
$D=-4( ad-bc)^{2}$
As given $ad\
eq bc$,
eq bc$,
$\therefore -4( ad-bc)^{2} < 0$
$\Rightarrow D< 0$
$\therefore$ Quadratic equation has no real roots.
Updated on: 10-Oct-2022
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