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ab = cd

ad = bc

${a^2} + {c^2} = {b^2} + {d^2}$

ac = bd

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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 $

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”.