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Question: If the equation\(({a^2} + {b^2}){x^2} - 2(ac + bd)x + ({c^2} + {d^2}) = 0\) has equal roots, then wh...

If the equation(a2+b2)x2−2(ac+bd)x+(c2+d2)=0({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
a2+c2=b2+d2{a^2} + {c^2} = {b^2} + {d^2}
ac = bd

Explanation

Solution

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 , (a2+b2)x2−2(ac+bd)x+(c2+d2)=0→(1)({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 Ax2+Bx+C=0A{x^2} + Bx + C = 0, equal roots are possible only when discriminant(D) = 0. Which can be found by the formula D=B2−4ACD = {B^2} - 4AC.
On comparing with equation 1, we get
A=(a2+b2) B=−2(ac+bd) C=(c2+d2) 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=[−2(ac+bd)]2−4×(a2+b2)×(c2+d2) =4(a2c2+b2d2+2ac⋅bd)−4(a2c2+b2d2+a2d2+b2c2)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=a2+b2+2ab{(a + b)^2} = {a^2} + {b^2} + 2ab
Now , D=0D = 0
⇒4(a2c2+b2d2+2ac⋅bd)−4(a2c2+b2d2+a2d2+b2c2)=0 ⇒(a2c2+b2d2+2ac⋅bd)=(a2c2+b2d2+a2d2+b2c2) ⇒a2d2+b2c2−2ac⋅bd=0 ⇒(ad−bc)2=0 ⇒ad−bc=0 ⇒ad=bc\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 Ax2+Bx+C=0A{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=B2−4ACD = {B^2} - 4AC . Now the three types of roots are as follows:
i). D>0 , Distinct and real roots exist and the roots are α=−B+D2A,β=−B−D2A\alpha = \dfrac{{ - B + \sqrt D }}{{2A}},\beta = \dfrac{{ - B - \sqrt D }}{{2A}}
ii). D=0, Real and equal roots exist and the roots are α=β=−B2A\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”.