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Solve that Show that (a² b²)x² 2(ac bd) x (c² d²) =0has no real root, if ad=not equal to bc.?
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Solve that Show that (a² b²)x² 2(ac bd) x (c² d²) =0has no real root, ...
Proof by Contradiction:

Assume that the given equation has a real root, say x = k.

Substituting k for x, we get:

(a² b²) k² + 2(ac bd) k + (c² d²) = 0

Using the quadratic formula, we get:

k = (-b ± √(b² - 4ac)) / 2a

Substituting the values of a, b, and c, we get:

k = (-2(ac bd) ± √((2(ac bd))² - 4(a² b²)(c² d²))) / 2(a² b²)

Simplifying this equation, we get:

k = (-1/ab) (ac bd ± √((ad - bc)²))

Since ad ≠ bc, we have:

(ad - bc)² > 0

Therefore, √((ad - bc)²) > 0

So, k cannot be real, as:

k = (-1/ab) (ac bd ± √((ad - bc)²))

and √((ad - bc)²) > 0

Hence, the given equation has no real root if ad ≠ bc.
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