MCQ
If $\left(a^2+b^2\right) x^2+2(a b+b d) x+c^2+d^2=0$ has no real roots, then :
- A$ab = bc$
- B$ab = cd$
- C$ac = bd$
- ✓$\text{ad}\neq\text{bc}$
✓
Answer
Correct option: D.
$\text{ad}\neq\text{bc}$
The given quadric equation is $\left(a^2+b^2\right) x^2+2(a b+b d) x+c^2+d^2=0$, and roots are equal.
Here, $a=\left(a^2+b^2\right), b=2(a b+b d)$ and, $c=c^2+d^2$
As we know that $D=b^2-4 a c$
Putting the value of $a=\left(a^2+b^2\right), b=2(a b+b d)$ and, $c=c^2+d^2$
$ =\{2(a b+b d)\}^2-4 \times\left(a^2+b^2\right) \times\left(c^2+d^2\right) $
$ =4 a^2 b^2+4 b^2 d^2+8 a b^2 d-4\left(a^2 c^2+a^2 d^2+b^2 c^2+b^2 d^2\right)$
$ =4 a^2 b^2+4 b^2 d^2+8 a b^2 d-4 a^2 c^2-4 a^2 d^2-4 b^2 c^2-4 a^2 d^2 $
$ =4 a^2 b^2+8 a b^2 d-4 a^2 c^2-4 a^2 d^2-4 b^2 c^2 $
$ =4\left(a^2 b^2+2 a b^2 d-a^2 c^2-a^2 d^2-b^2 c^2\right)$
The given equation will have no real roots, if $ {D}<0$
$4\left(a^2 b^2+2 a b^2 d-a^2 c^2-a^2 d^2-b^2 c^2\right)<0 $
$ a^2 b^2+2 a b^2 d-a^2 c^2-a^2 d^2-b^2 c^2<0$
$\text{ad}\neq\text{bc}$
Thus, the correct answer is $(d)$
Here, $a=\left(a^2+b^2\right), b=2(a b+b d)$ and, $c=c^2+d^2$
As we know that $D=b^2-4 a c$
Putting the value of $a=\left(a^2+b^2\right), b=2(a b+b d)$ and, $c=c^2+d^2$
$ =\{2(a b+b d)\}^2-4 \times\left(a^2+b^2\right) \times\left(c^2+d^2\right) $
$ =4 a^2 b^2+4 b^2 d^2+8 a b^2 d-4\left(a^2 c^2+a^2 d^2+b^2 c^2+b^2 d^2\right)$
$ =4 a^2 b^2+4 b^2 d^2+8 a b^2 d-4 a^2 c^2-4 a^2 d^2-4 b^2 c^2-4 a^2 d^2 $
$ =4 a^2 b^2+8 a b^2 d-4 a^2 c^2-4 a^2 d^2-4 b^2 c^2 $
$ =4\left(a^2 b^2+2 a b^2 d-a^2 c^2-a^2 d^2-b^2 c^2\right)$
The given equation will have no real roots, if $ {D}<0$
$4\left(a^2 b^2+2 a b^2 d-a^2 c^2-a^2 d^2-b^2 c^2\right)<0 $
$ a^2 b^2+2 a b^2 d-a^2 c^2-a^2 d^2-b^2 c^2<0$
$\text{ad}\neq\text{bc}$
Thus, the correct answer is $(d)$
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