MCQ
If $\left(a^2+b^2\right) x^2+2(a c+b d) x+c^2+d^2=0$ has no real roots, then
- A$ab = cd$
- B$ad=bc$
- ✓$ad \neq bc$
- D$ac = bd$
✓
Answer
Correct option: C.
$ad \neq bc$
$\left( a ^2+ b ^2\right) x ^2+2( ac + bd ) x + c ^2+ d ^2=0$
Here $A=a^2+b^2, B=2(a c+b d), C=c^2+d^2$
$D=B^2-4 A C=[2(a c+b d)]^2-4\left(a^2+b^2\right)\left(c^2+d^2\right)$
$=4\left[a^2 c^2+b^2 d^2+2 a b c d\right]-4\left[a^2 c^2+a^2 d^2+b^2 c^2+b^2 d^2\right]$
$=4 a^2 c^2+4 b^2 d^2+8 a b c d-4 a^2 c^2-4 a^2 d^2-4 b^2 c^2-4 b^2 d^2$
$=-4 a^2 d^2-4 b^2 c^2+8 a b c d$
$=-4\left(a^2 d^2+b^2 c^2-2 a b c d\right)$
$=-4(a d-b c)^2$
$\because$ Roots are not real
$\therefore D<0$
$\therefore-4(a d-b c)^2<0 \Rightarrow(a d-b c)^2<0$
$\Rightarrow a d-b c<0 \text { or } a d \neq b c$
Here $A=a^2+b^2, B=2(a c+b d), C=c^2+d^2$
$D=B^2-4 A C=[2(a c+b d)]^2-4\left(a^2+b^2\right)\left(c^2+d^2\right)$
$=4\left[a^2 c^2+b^2 d^2+2 a b c d\right]-4\left[a^2 c^2+a^2 d^2+b^2 c^2+b^2 d^2\right]$
$=4 a^2 c^2+4 b^2 d^2+8 a b c d-4 a^2 c^2-4 a^2 d^2-4 b^2 c^2-4 b^2 d^2$
$=-4 a^2 d^2-4 b^2 c^2+8 a b c d$
$=-4\left(a^2 d^2+b^2 c^2-2 a b c d\right)$
$=-4(a d-b c)^2$
$\because$ Roots are not real
$\therefore D<0$
$\therefore-4(a d-b c)^2<0 \Rightarrow(a d-b c)^2<0$
$\Rightarrow a d-b c<0 \text { or } a d \neq b c$
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