Answer :

Compare the given quadratic equation with ax2 + bx + c = 0

Here,


The coefficient of x2 is a


The coefficient of xis b


And, c is the intercept.


Now, here:


a = a2 + b2


b = 2(ac + bd)


c = c2 + d2


Consider,


Then,


b2 – 4ac = [2(ac + bd)]2 – 4(a2 + b2)(c2 +d2)


= 4[a2c2 + 2abcd + b2d2] – 4[a2c2 + a2d2 + b2c2 + b2d2]


= 4a2c2 + 8abcd + 4b2d2 – 4a2c2 – 4a2d2 – 4b2c2 – 4b2d2


= 8abcd– 4a2d2 – 4b2c2


= –4[4a2d2 + 4b2c2 – 2abcd]


= –4[ad – bc]2


Hence the given equation has no real roots unless ad = bc


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