Question

what to do after this
I am struck..
the ans is incomplete and plz help me...

Answer 1

Hey There!



Here the equation is:

$x^2(a^2+b^2)+2(ac+bd)x+(c^2+d^2)=0$


Let us first find the Discriminant:

$D=b^2-4ac \\ \\ \implies D=4(ac+bd)^2-4(a^2+b^2)(c^2+d^2) \\ \\ \implies D = 4((a^2c^2+2abcd+b^2d^2)-(a^2c^2+a^2d^2+b^2c^2+b^2d^2)) \\ \\ \implies D = 4(a^2c^2+2abcd+b^2d^2-a^2c^2-a^2d^2-b^2c^2-b^2d^2) \\ \\ \implies D= 4(2abcd-a^2d^2-b^2c^2) \\ \\ \implies D=-4(a^2d^2-2(ad)(bc)+b^2c^2) \\ \\ \implies D=-4(ad-bc)^2$



Now, here, we are given that $ad\neq bc$

This means that  $(ad-bc)^2 \neq 0$ 

Thus, Discriminant cannot be zero.

Now, we know that a perfect square is always positive.\

We have:

$(ad-bc)^2 > 0 \\ \\ \implies -4(ad-bc)^2 < 0 \\ \\ D<0$


Thus, The quadratic equation has no real roots, provided $ad\neq bc$

Hence Proved.

Hope it helps
Purva
Brainly Community

Answer 2

Hey mate!

Here's your answer!!

Kindly refer to the given attachment.

$hope \: it \: helps \: you...$
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#BE BRAINLY