Question

If ad ≠ bc, then prove that the equation,


(a² + b²
) x² + 2 (ac + bd) x + (c² + d²
) = 0 has no real roots.

​

Answer 1

Answer:

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Answer 2

Given:-

ab≠bc

To prove :-

$( {a}^{2} + {b}^{2} ) {x}^{2} + 2(ac + bd)x + ( {c}^{2} + {d}^{2} ) < 0$

According to quadratic equation (ax²+bx+c) has some defined formulae for the discriminant (D) ;

${b}^{2} - 4ac = 0 \: \\\:(real \: and \: equal \: roots)$

${b}^{2} - 4ac < 0 \\ (no \: real \: roots)$

${b}^{2} - 4ac > 0 \\ (real \: and \: distinct \: roots)$

As the given equation has no real roots so we will use the D for no real roots,

Then the values are :

$a \: \: = \: ( {a}^{2} + {b}^{2} \: )$

$b \: \: = 2(ac + bd)$

$c \: \: = ( {c}^{2} + {d}^{2} \: )$

Now on putting these values we get,

$= > {b}^{2} - 4ac < 0$

$= > {[2(ac + bd)] }^{2} - \\ 4( {a}^{2} + {b}^{2} )( {c}^{2} + {d}^{2} ) < 0$

$= >[ 4( {a}^{2} {c}^{2} + 2acbd + {b}^{2} {d}^{2} ) \\ - 4( {a}^{2} + {b}^{2} )( {c}^{2} + {d}^{2} )] < 0$

$= > [4({a}^{2} {c}^{2} + 2acbd + {b}^{2} {d)}^{2} \\ - 4( {a}^{2} {c}^{2} + {a}^{2} {d}^{2} + {b}^{2} {c}^{2} + {b}^{2} {d}^{2} )] < 0$

$=>[4({a}^2{c}^2+2acbd+{b}^2{d}^2)\\-4( {a}^{2} {c}^{2} + {a}^{2} {d}^{2} + {b}^{2} {c}^{2} + {b}^{2} {d}^{2}])] <0$

$=>[-4{a}^2{d}^2+8acbd-4{b}^2{c}^2]<0$

$=>[{a} ^2{d}^2-2acbd+{b}^2{c}^2] <0$

$=>{(ad-bc)} ^2<0$

$=>(ad-bc) <0$

$=>(ad≠bc)$

Hence proved ad≠bc and the equation has no real roots.