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If the equation$({a^2} + {b^2}){x^2} - 2(ac + bd)x + ({c^2} + {d^2}) = 0$ has equal roots, then which one of the following is correct?
ab = cd
ad = bc
${a^2} + {c^2} = {b^2} + {d^2}$
ac = bd

Answer
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Hint: Start by comparing with the standard quadratic equation and find out the discriminant value, and according to the condition of equal roots equate D to zero i.e. D=0, Find out the required condition by simplifying the relation.

Complete step-by-step solution:
Given , $({a^2} + {b^2}){x^2} - 2(ac + bd)x + ({c^2} + {d^2}) = 0 \to (1)$
Step by step solution
We know for any quadratic equation $A{x^2} + Bx + C = 0$, equal roots are possible only when discriminant(D) = 0. Which can be found by the formula $D = {B^2} - 4AC$.
On comparing with equation 1, we get
$
  A = ({a^2} + {b^2}) \\
  B = - 2(ac + bd) \\
  C = ({c^2} + {d^2}) $
Using the above concept we need to find discriminant (D), we get :
$ D = {\left[ { - 2(ac + bd)} \right]^2} - 4 \times ({a^2} + {b^2}) \times ({c^2} + {d^2}) \\
   = 4({a^2}{c^2} + {b^2}{d^2} + 2ac \cdot bd) - 4({a^2}{c^2} + {b^2}{d^2} + {a^2}{d^2} + {b^2}{c^2}) $
Here, we used the formula ${(a + b)^2} = {a^2} + {b^2} + 2ab$
Now , $D = 0$
$\Rightarrow 4({a^2}{c^2} + {b^2}{d^2} + 2ac \cdot bd) - 4({a^2}{c^2} + {b^2}{d^2} + {a^2}{d^2} + {b^2}{c^2}) = 0 \\
   \Rightarrow ({a^2}{c^2} + {b^2}{d^2} + 2ac \cdot bd) = ({a^2}{c^2} + {b^2}{d^2} + {a^2}{d^2} + {b^2}{c^2}) \\
   \Rightarrow {a^2}{d^2} + {b^2}{c^2} - 2ac \cdot bd = 0 \\
   \Rightarrow {(ad - bc)^2} = 0 \\
   \Rightarrow ad - bc = 0 \\
   \Rightarrow ad = bc $
So, option B is the correct option.

Note: For any quadratic equation $A{x^2} + Bx + C = 0$, there are three types of roots available which can only be determined after calculating Discriminant(D) by the formula or relation $D = {B^2} - 4AC$ . Now the three types of roots are as follows:
i). D>0 , Distinct and real roots exist and the roots are $\alpha = \dfrac{{ - B + \sqrt D }}{{2A}},\beta = \dfrac{{ - B - \sqrt D }}{{2A}}$
ii). D=0, Real and equal roots exist and the roots are $\alpha = \beta = \dfrac{{ - B}}{{2A}}$
iii). D<0, Imaginary roots exist.
For imaginary roots, there's a whole different chapter and concept known as “Complex numbers and roots”.