Question

If ad is not equal to bc, then prove that the equation

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

Answer 1

HELLO DEAR,


Hence the given equation has no real roots unless AD not equal BC

I HOPE ITS HELP YOU DEAR,
THANKS

Answer 2

"Given:

$\left(a^{2}+b^{2}\right) x^{2}+2(a c+b d) x+\left(c^{2}+d^{2}\right)=0$

To prove:

$\left(a^{2}+b^{2}\right) x^{2}+2(a c+b d) x+\left(c^{2}+d^{2}\right)=0$ has no real roots.

Solution:

Let us check the given equation with $A x^{2}+B x+C=0$

Here, $A=a^{2}+b^{2}$

B = 2(ac + bd)

$C=c^{2}+d^{2}$

Consider, $B^{2}-4 A C=[2(a c+b d)]^{2}-4\left(a^{2}+b^{2}\right) \times\left(c^{2}+d^{2}\right)$

$=4\left[a^{2} c^{2}+2 a b c d+b^{2} d^{2}\right]-4\left[a^{2} c^{2}+a^{2} d^{2}+b^{2} c^{2}+b^{2} d^{2}\right]$

$=4 a^{2} c^{2}+8 a b c d+4 b^{2} d^{2}-4 a^{2} c^{2}-4 a^{2} d^{2}-4 b^{2} c^{2}-4 b^{2} d^{2}$

$=8 a b c d-4 a^{2} d^{2}-4 b^{2} c^{2}$

$=-4\left[a^{2} d^{2}+b^{2} c^{2}-2 a b c d\right]$

$B^{2}-4 A C =-4[a d-b c]^{2}$

If ad = bc, then we get, $\left(a^{2}+b^{2}\right) x^{2}+2(a c+b d) x+\left(c^{2}+d^{2}\right)=0$

Thus, the given equation has real roots when ad = bc and the given equation has real roots when ad and bc are not equal."